# What is the surface area of the solid created by revolving #f(x) = (x-9)^2 , x in [2,3]# around the x axis?

These are two known integrals. They can be found using iterative integration by parts and are quite cumbersome. I recommend looking them up or committing them to memory, since they're fairly common in trigonometric substitution questions:

Then:

Thus:

Moving through this quickly:

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To find the surface area of the solid created by revolving the function ( f(x) = (x-9)^2 ) over the interval ([2,3]) around the x-axis, you can use the formula for the surface area of a solid of revolution:

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

where ( y = f(x) ), and ( a ) and ( b ) are the limits of integration.

Evaluate ( \frac{{dy}}{{dx}} ), which is the derivative of ( f(x) ). Then, compute ( 2\pi y \sqrt{1 + \left(\frac{{dy}}{{dx}}\right)^2} ) for each ( x ) in the interval ([2,3]). Finally, integrate this expression from 2 to 3 with respect to ( x ) to find the surface area of the solid.

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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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