Volume of a solid bounded by functions Find the volume of a solid bounded (sup) by $z=\sqrt{4-x^2-y^2}$ and (inf) by $r=2\cos(\theta)$
So i guess the integral of the volume can be expresses by: $$\int_{0}^{2\pi}\int_{0}^{2\cos(\theta)}\int_{0}^{\sqrt{4-r^2}}1 dzdrd\theta$$
but the solution of this integral is equal to 0. However the solution of the problem is $$\frac{8\pi}{3}-\frac{32}{9}$$
 A: Hint:
I suppose that you are using cylindrical coordinates, so $z=\sqrt{4-x^2-y^2}$ is the half-sphere ($z\ge 0$) centered at the origin and of radius $r=2$. And  $r=2\cos(\theta)$ is the cylinder with axis parallel to the $z$ axis and centered at $(1,0,0)$ and radius $ R = 1$ (so it is a bit confusing to speak of sup and inf  bounds).
So what you want is the volume of a ''cylindrical hole'' in a sphere that has the diameter equal to the radius of the sphere, and this is a classical problem that is solved as you can see here: Volume of off-center cylindrical hole in Sphere
If you look at the answer to this question you can see that your integral has two misteakes: the first is the volume element in cylindrical coordinates that is $dV=r dz dr d\theta$, the second is in the limits of integration.
The better way is to use the symmetry of the hole around the  $x-z$ axis, and evaluate the volume as:
$$
V=2\int_0^{\frac{\pi}{2}}\int_{2\cos \theta}^0\int_0^{\sqrt{4-r^2}}rdzdrd\theta =2\int_0^{\frac{\pi}{2}}\int_{2\cos \theta}^0r\sqrt{4-r^2}drd\theta =
$$ 
$$
=\frac{2}{3}\int_0^{\frac{\pi}{2}}8\cdot(1+\sin^3\theta)d\theta=\frac{16}{3}\left( \frac{\pi}{2}-\frac{2}{3}\right)
$$
A: You're very close, here a picture of $r=2\cos(\theta)$.

This is expected as $r^2=x^2+y^2=2r\cos(\theta)=2x$. So we have $(x-1)^2+y^2=1$
Notice however that $\theta \in [-\frac{\pi}{2},\frac{\pi}{2}]$. 
One last thing don't forget your jacobian, $dxdydz=r dz dr d \theta $. 
We have,
$$V=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos(\theta)}\int_{0}^{\sqrt{4-r^2}}r dzdrd\theta$$
