# Help with finding the surface area of a revolution

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.

-

\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}
@OghmaOsiris: The calculation that led from your displayed equation $2$ to displayed equation $3$ has a minor algebra error. Probably did not notice that $(-1)^2=1$. – André Nicolas Sep 1 '11 at 16:26
@OghmaOsiris: How you got $9 - y^2 -y^2$ is a complete mystery to me. But I can say that that's where you went wrong. If I knew how you did it, maybe I could be more specific. – Michael Hardy Sep 1 '11 at 16:35
@AndréNicolas That's it. I redid the problem and I was able to come to the right conclusion after I redid $(-y)^2 = y^2$. – OghmaOsiris Sep 1 '11 at 16:49