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

Let's momentarily leave out the bounds and return to the bounds once we complete integration.

Unfortunately, these integrals are both quite messy, so attached are links to finding both the integrals:

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

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

First, find ( f'(x) ) by differentiating ( f(x) ):

[ f'(x) = 4(2x - 2) ]

Then, plug ( f(x) ) and ( f'(x) ) into the surface area formula:

[ SA = \int_{1}^{2} 2\pi (2x - 2)^2 \sqrt{1 + (4(2x - 2))^2} , dx ]

Compute the integral to find the surface area.

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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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