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Here's the problem I'm stuck on:

Find the surface are of this revolution about the y-axis

$x = \sqrt{9-y^2}; -2\leq y\leq2$

What I've done so far:

$A= 2\pi \int_{-2}^2 \sqrt{9-y^2} \sqrt{1 + (\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y))^2} dy$

$ = 4\pi \int_{0}^2 \sqrt{9-y^2} \sqrt{1 + (\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y))^2} dy$

$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)(\frac{1}{2}(-2y)(9-y^2)^\frac{-1}{2})^2} dy$

$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)((-y)^2(9-y^2)^{-1})} dy$

$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (9-y^2)(\frac{(-y)^2}{(9-y^2)})} dy$

$ = 4\pi \int_{0}^2 \sqrt{(9-y^2) + (-y)^2} dy$

$ = 4\pi \int_{0}^2 \sqrt{9-y^2 -y^2} dy$*

$ = 4\pi \int_{0}^2 \sqrt{9-2y^2} dy$

The answer in the book says its:

$24\pi$

Which means that I needed to get the integral to be:

$ = 4\pi \int_{0}^2 \sqrt{9} dy$

But I just don't see how I can manipulate the problem with algebra to get there... Any guidance?

EDIT:

Added in some steps to show where my algrbra went wrong

*This is where I made the mistake.

1 Answers 1

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$ \begin{align} & \sqrt{9-y^2} \sqrt{1 + \left(\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y)\right)^2} = \sqrt{9-y^2} \sqrt{1 + \frac{4y^2}{4(9-y^2)}} \\ & = \sqrt{9-y^2} \sqrt{1 + \frac{y^2}{9-y^2}} = \sqrt{9-y^2} \sqrt{\frac{(9-y^2) + y^2}{9-y^2}} \\ & = \sqrt{9-y^2} \sqrt{\frac{9}{9-y^2}} = \sqrt{9}. \end{align} $

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    So you had $9-y^2 + (-y)^2$. That's the same as $9-y^2+y^2$, which simplifies to $9$.2011-09-01