<s>[INST] 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.
 [/INST] $$
\begin{align}
& \sqrt{9-y^2} \sqrt{1 + \left(\frac{1}{2}(9-y^2)^\frac{-1}{2}(-2y)\right)^2}
= \sqrt{9-y^2} \sqrt{1 + \frac{4y^2}{4(9-y^2)}} \\
& = \sqrt{9-y^2} \sqrt{1 + \frac{y^2}{9-y^2}}
= \sqrt{9-y^2} \sqrt{\frac{(9-y^2) + y^2}{9-y^2}} \\
& = \sqrt{9-y^2} \sqrt{\frac{9}{9-y^2}} = \sqrt{9}.
\end{align}
$$
</s>