# Surface area of the part of the sphere $x^2+y^2+z^2=a^2$ that is inside the cylinder $x^2+y^2=ax$

I've been solving some surface area problems lately, but I don't think that the same approach that I was using will work with this one (or at least will result in a lot work).

So, I believe I should use spherical coordinates for the parameterization, I did:

$$x=r\sin\phi\cos\theta, y=r\sin\phi\sin\theta, z=r\cos\phi$$

When I take the absolute value (is the term "absolute value" correct?) of the cross product of the partial derivatives of $r(\rho, \space \theta)$ I will get $\sin\phi$ (done it before). Therefore the integral I have to evaluate in order to get what is asked is:

$$\iint_D\sin\phi \space d_\theta d_\phi$$

I'm just not sure of how to define the boundaries for $\theta$. What I did on other problems was define $R(r, \space \theta) = (r\cos\theta, r\sin\theta, z)$ (the cylinder parameterization) where $z$ would be the surface, will this work here? I have a feeling that taking the cross product of something like $\sqrt{a^2-x^2-y^2}$ will be pretty hairy and a more elegant way of finding $\theta$ should exist. If anyone can point out what I'm not seeing here I will be very glad. Thank you.

You definitely want to do this with a polar coordinate parametrization of the sphere, as then you're parametrizing (half) the surface you're interested in by $(r,\theta)$, with $-\pi/2\le\theta\le\pi/2$ and $0\le r\le a\cos\theta$. The integrand is then $d\sigma = \dfrac{dA}{\cos\gamma}$, where $\gamma$ is the angle between the vertical and the surface normal. This comes out quite nicely in polar coordinates.
EDIT: If you do insist, you can do it, however, in spherical coordinates. You're missing a factor of $a^2$ in your integral. Then note that the equation of the cylinder becomes $a^2\sin^2\phi = a^2\sin\phi\cos\theta$, i.e., $\sin\phi = \cos\theta$, so $-\pi/2\le\theta\le\pi/2$, and $0\le \phi\le\arcsin(\cos\theta)$. This is the same integral you end up with using the polar coordinates parametrization.