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

x = \sqrt{a^2 - y^2} on 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 \sqrt{2}, like this:

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

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

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

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

**Is this right?

So my integral is:

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

Then I pulled my square root out in front:

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

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