I'm working on studying for the GRE. I did this problem from Stewart's Calculus, but my answer differs from that in the back of the book.
The problem is: Find the area of the part of the sphere $x^{2} + y^{2} + z^{2} = a^{2}$ that lies within the cylinder $x^{2} + y^{2} = ax$ and above the $xy$-plane.
I let $\varphi(x,y) = (x,y,\sqrt{a^{2} - x^{2} - y^{2}})$ be the coordinate function describing the upper half of the sphere. A routine computation gives $$\left \|\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y} \right \| = \frac{a}{\sqrt{a^{2} - (x^{2} + y^{2})}}$$
We integrate over the circle $x^{2} + y^{2} = ax$, which can be parametrized in polar coordinates as $\frac{-\pi}{2} \leq \theta \leq \frac{\pi}{2}, \, \, 0 \leq r \leq a \cos(\theta)$. Integration gives \begin{align*} SA &= \frac{-a}{2} \int_{- \pi/2}^{\pi/2} \int_{0}^{a \cos(\theta)} \frac{-2r}{\sqrt{a^{2} - r^{2}}} \, dr d\theta \\ &= \frac{-a}{2} \int_{-\pi/2}^{\pi/2} \left( 2 \sqrt{a^{2} - a^{2} \cos^{2}(\theta)} - 2a \right) \, d\theta \\ &= a^{2} \int_{-\pi/2}^{\pi/2} \, d\theta - a^{2} \int_{-\pi/2}^{\pi/2} \sin(\theta) \, d\theta \end{align*} which gives me $SA = \pi a^{2}$, but the book says $SA = a^{2}(\pi - 2)$. I can see how that would have been the right answer if I had gotten a cosine in the last integral instead of a sine, but it's not clear to me where I went wrong.
The problem is $$\sqrt{1-\cos^2\theta}$$ in $[-\frac{\pi}{2},-\frac{\pi}{2}]$ has two different values. It is $-\sin\theta$ on $[-\frac{\pi}{2},0]$ and $\sin\theta$ on $[0,\frac{\pi}{2}]$. So you need to integrate them separately.