# What is the surface area of the solid created by revolving #f(x)=x-1# for #x in [1,2]# around the x-axis?

Integrating over the interval, the surface is then calculated as:

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The surface area of the solid created by revolving ( f(x) = x - 1 ) for ( x ) in ([1,2]) around the x-axis can be calculated using the formula for surface area of revolution:

[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

where ( a ) and ( b ) are the limits of integration, and ( \frac{dy}{dx} ) is the derivative of ( f(x) ) with respect to ( x ).

In this case, ( a = 1 ), ( b = 2 ), and ( f(x) = x - 1 ). The derivative of ( f(x) ) is ( \frac{dy}{dx} = 1 ).

Substitute these values into the formula:

[ S = 2\pi \int_{1}^{2} (x - 1) \sqrt{1 + (1)^2} , dx ]

[ S = 2\pi \int_{1}^{2} (x - 1) \sqrt{2} , dx ]

[ S = 2\pi \sqrt{2} \int_{1}^{2} (x - 1) , dx ]

[ S = 2\pi \sqrt{2} \left[ \frac{x^2}{2} - x \right]_{1}^{2} ]

[ S = 2\pi \sqrt{2} \left[ \frac{2^2}{2} - 2 - \left( \frac{1^2}{2} - 1 \right) \right] ]

[ S = 2\pi \sqrt{2} \left[ 2 - 2 - \frac{1}{2} + 1 \right] ]

[ S = 2\pi \sqrt{2} \left[ - \frac{1}{2} + 1 \right] ]

[ S = 2\pi \sqrt{2} \left[ \frac{1}{2} \right] ]

[ S = \pi \sqrt{2} ]

Therefore, the surface area of the solid is ( \pi \sqrt{2} ) square units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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