Surface parametrization and area

Question

I am trying to parametrize and find the area of the portion of the sphere $x^2 + y^2 + z^2 = a^2$ inside $x^2+y^2=ax$, with $a>0$. Firstly, I tried to find the intersection $ax+z^2=a^2$, but I'm not sure whether this helps or not.

Answer 1

$x^2 + y^2 + z^2 = a^2$ suggests:

$x = a \cos\theta \sin\phi\\ y = a \sin\theta \sin\phi\\ z = a \cos\phi\\ $

So what does the other constraint do?

$a^2 \sin^2 \phi = a^2 \cos\theta \sin \phi\\ \sin \phi = \cos\theta$

This is the seam where the surfaces intersect. The surface of the sphere inside the cylinder is:

$\sin \phi\le \cos\theta$

But it suggests that this would be a little bit more elegant.

$x = a \sin\theta \sin\phi\\ y = a \cos\theta \sin\phi\\ z = a \cos\phi$

$\phi \le \theta$ when $0 \le \theta < \pi/2$

$\phi \le \pi-\theta$ when $\pi/2 \le \theta < \pi$

and

$\phi \ge \pi-\theta$ when $0 \le \theta < \pi/2$

$\phi \ge \theta$ when $\pi/2 \le \theta < \pi$

And $0\le\theta<\pi$,because the entire cylinder is on one side of the x axis. And once we have swept out $\pi$ radians, we have traced out the entire cylinder.