This was on my exam, and I didn't know how to solve it.
Problem: Find the area of the sphere $x^2 + y^2 +z^2 = 4$ inside the cylinder $x^2 + y^2 = 2x$ above the $xy$-plane.
The way is tried to do it was that $dS=\sqrt{1+(dz/dx)^2 + (dz/dy)^2 } dA $
But when I tried to integrate I got a really messy integral.
Any help is appreciated.
The surface integral is,
$$S = \int_{A} \sqrt{1+(z_x')^2+(z_y')^2}dA =\int_{A} \frac2{\sqrt{4-x^2-y^2}}dA$$
Then, evaluate the integral in polar coordinates of the cylinder $r=2\cos\theta$,
$$S = 2\int_{-\pi/2}^{\pi/2} d\theta\int_0^{2\cos\theta} \frac{rdr}{\sqrt{4-r^2}} =-4\int_{0}^{\pi/2} \sqrt{4-r^2}|_0^{2\cos\theta}d\theta$$
$$ =8\int_{0}^{\pi/2} (1-\sin\theta)d\theta=4(\pi-2)$$