Find the volume common to the surfaces $y^2+z^2=4ax$ and $x^2+y^2=2ax$
I set up the following integral :
$$\iiint_{z=-\sqrt{4ax-y^2}}^{\sqrt{4ax-y^2}}\,dz\,dy\,dx = 2\cdot \int_{x=0}^{2a} \int_{y=-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}} \sqrt{4ax-y^2}\,dy\,dx$$
Which is not easy to integrate.
If I use polar coordinates the integration becomes
$$2\cdot \int_{\theta=-\pi/2}^{\pi/2} \int_{r=0}^{2a\cos(\theta)} \sqrt{4ar\cos(\theta)-r^2\sin^2(\theta)} \cdot r\,dr\,d\theta$$
Which is again difficult