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

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To find the surface area of the solid created by revolving the function ( f(x) = \sqrt{x} ) for ( x ) in the interval ([1,2]) around the x-axis, you can use the formula for surface area of a solid of revolution:

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

Where ( f'(x) ) represents the derivative of ( f(x) ).

In this case, ( f(x) = \sqrt{x} ). The derivative ( f'(x) = \frac{1}{2\sqrt{x}} ).

So,

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

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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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