What is the surface area produced by rotating #f(x)=1/(x+1), x in [0,3]# around the x-axis?

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surface area of a curve rotated around the xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axisTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ ATo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ STo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{aTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\piTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{bTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_aTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^bTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\piTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrtTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrtTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} ,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dxTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} ,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

FirstTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where: To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we findTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where: -To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( aTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) andTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( bTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) areTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limitsTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits ofTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration. To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration. -To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) isTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the functionTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representingTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curveTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve. To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) isTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivativeTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative ofTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

NowTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

FirstTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, weTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plugTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, weTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) intoTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surfaceTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface areaTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula andTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrateTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the intervalTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3]To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ STo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\piTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

NowTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substituteTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrtTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) )To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) andTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and (To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( fTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \leftTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x)To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) intoTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surfaceTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface areaTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formulaTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ ATo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\rightTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} ,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\piTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrtTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\rightTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} ,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} ,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

ThisTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ ATo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integralTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A =To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral mayTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may notTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \intTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not haveTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have aTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simpleTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closedTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-formTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solutionTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution.To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. ThereforeTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore,To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\piTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methodsTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \fracTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximationTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniquesTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, suchTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such asTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as SimpsonTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson'sTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's ruleTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numericalTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrtTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integrationTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithmsTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, canTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can beTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be usedTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximateTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate theTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate the surface areaTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(xTo find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate the surface area.To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(x +To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate the surface area.To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(x + To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate the surface area.To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(x + 1To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) around the x-axis over the interval ( x ) in ([0, 3]), we use the formula for the surface area of a curve rotated around the x-axis:

[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} , dx ]

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ] [ f'(x) = -\frac{1}{(x + 1)^2} ]

Now, we plug ( f(x) ) and ( f'(x) ) into the surface area formula and integrate over the interval ( [0, 3] ):

[ S = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{(x + 1)^2}\right)^2} , dx ]

[ = 2\pi \int_0^3 \frac{1}{x + 1} \sqrt{1 + \frac{1}{(x + 1)^4}} , dx ]

This integral may not have a simple closed-form solution. Therefore, numerical methods or approximation techniques, such as Simpson's rule or numerical integration algorithms, can be used to approximate the surface area.To find the surface area produced by rotating ( f(x) = \frac{1}{x + 1} ) on the interval ( x \in [0, 3] ) around the x-axis, we use the formula for the surface area of a solid of revolution:

[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + (f'(x))^2} , dx ]

Where:

• ( a ) and ( b ) are the limits of integration.
• ( f(x) ) is the function representing the curve.
• ( f'(x) ) is the derivative of ( f(x) ).

First, we find ( f'(x) ):

[ f(x) = \frac{1}{x + 1} ]

[ f'(x) = -\frac{1}{{(x + 1)}^2} ]

Now, we substitute ( f(x) ) and ( f'(x) ) into the surface area formula:

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \left(-\frac{1}{{(x + 1)}^2}\right)^2} , dx ]

[ A = \int_{0}^{3} 2\pi \frac{1}{x + 1} \sqrt{1 + \frac{1}{{(x + 1)}^4}} , dx ]

This integral may not have a simple closed-form solution. Numerical methods or approximation techniques may be used to compute the value of this integral.