Find surface area of solid formed by rotating the following curve about the y-axis:

$\displaystyle x = \sqrt{a^2 - y^2}$ on $\displaystyle 0 \leq y \leq \frac{a}{2}$

*The "a" in there kinda screwed me up, but this is my work:

When I solved for the arc length, I ended up getting $\displaystyle \sqrt{2}$, like this:

$\displaystyle x = \sqrt{a^2 - y^2}$

$\displaystyle x' = \frac{2a - 2y}{2\sqrt{a^2-y^2}}$

$\displaystyle = \frac{a-y}{a-y} = 1$

$\displaystyle ds = \sqrt{1 + (1)^2}$

**Is this right?

So my integral is:

$\displaystyle \int\limits^{\frac{a}{2}}_{0} 2 \pi (\sqrt{a^2 - y^2}) (\sqrt{2})$

Then I pulled my square root out in front:

$\displaystyle 2\sqrt{2} \pi \int\limits^{\frac{a}{2}}_{0} \sqrt{a^2 - y^2} dy$

$\displaystyle 2\sqrt{2} \pi \int\limits^{\frac{\pi}{6}}_{0} asin \theta \cdot acos \theta d\theta$

I have more, but can some one check this up until this point?
Thanks!! Molly