Computing volume inside a ball and outside a cylinder I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral. 
Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form and so I get
$$\ x^2 + y^2 + z ^2 = 4 \Rightarrow z = \sqrt{4-r^2} \\x^2 + y^2 = 2x \Rightarrow r = 2 \cos \theta $$
therefore the integral should be 
$$\ 4 \cdot \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$$
but this integral is way to messy for me to calculate. I mean first step of calculating integral for $\ r $ is okay but then the value I get and calculating integral of $\ \theta $ is beyond me and I believe it is beyond the scope of the course. I guess I'm missing something in the process here?

Let $I =  \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$
$I = \frac{-1}{2}\int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2})  \ d(4-r^2) \ d \theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-r^2)^{3/2}\big]^{2cos\theta}_0d\theta$
$I = \frac{-1}{2}.\frac{2}{3}\int_0^{\pi/2}\big[(4-4cos^2\theta)^{3/2} - 4^{3/2}\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^2\theta)^{3/2} - 1\big]d\theta$
$I = \frac{-1}{3}.2^3\int_0^{\pi/2}\big[(sin^3\theta)- 1\big]d\theta$
$\bigg[ \int^{\pi/2}_0sin^mxdx = \frac{m-1}{m}.\frac{m-3}{m-2}...1 \bigg]$, if m is odd
$I = \frac{-2^3}{3}\bigg[\big[\frac{2}{3}.1\big] -\frac{\pi}{2}\bigg]$
Thus, $I = \frac{8}{18}[3\pi - 4] = \frac{4}{9}[3\pi - 4]$ 
Can you proceed now?
