Surface area of sphere $x^2 + y^2 + z^2 = a^2$ cut by cylinder $x^2 + y^2 = ay$, $a>0$ The cylinder is given by the equation $x^2 + (y-\frac{a}{2})^2 = (\frac{a}{2})^2$.
The region of the cylinder is given by the limits $0 \le \theta \le \pi$, $0 \le r \le a\sin \theta$ in polar coordinates.
We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:
$$A=2\iint\frac{\sqrt{\left(\frac{\partial F}{\partial x}\right)^2 + \left(\frac{\partial F}{\partial y}\right)^2 + \left(\frac{\partial F}{\partial z}\right)^2}}{\left|\frac{\partial F}{\partial z} \right|} dA$$
where $F$ is the equation of the sphere.
Plugging in the expressions and simplifying ($z \ge 0)$, we get:
$$A=2a\iint\frac{1}{\sqrt{a^2 - x^2 - y^2}} dxdy$$
Converting to polar coordinates, we have:
$$A = 2a \int_{0}^\pi \int_{0}^{a\sin(\theta)} \frac{r}{\sqrt{a^2 - r^2}} drd\theta$$
Calculating this I get $2\pi a^2$. The answer is $(2\pi - 4)a^2$. Where am I  going wrong?

I know what you r doing rong, I solved this a week ago, the same way you did.
You forgot that $\sqrt{\sin^2\theta} = |\sin\theta|$, not $\sin\theta$ 
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This is how you might have done 
$$ 2\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}\dfrac{a}{\sqrt{a^2-r^2}}\cdot r \cdot  drd\theta$$
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$$= 2\int_{-\pi/2}^{\pi/2}\left[ -a\sqrt{a^2-r^2}\right]_0^{a\cos\theta}d\theta$$
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$$= 2\int_{-\pi/2}^{\pi/2}- a^2\sqrt{\sin^2\theta}-\left(-a^2 \right)d\theta$$
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$$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2\sin\theta d\theta$$
MISTAKE !!!!
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INSTEAD  $$= 2\int_{-\pi/2}^{\pi/2}a^2- a^2|\sin\theta| d\theta$$
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$$= 4\int_{0}^{\pi/2}a^2- a^2\sin\theta d\theta$$
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$$= 4a^2 \left[1-  \sin\theta \right] _{0}^{\pi/2}$$
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$$= a^2\left(2\pi-4 \right)$$
