Find the volume of the region bounded by a sphere and a paraboloid using cylindrical and spherical coordinates. The sphere is $x^2 + y^2 + z^2 = 4$ and the paraboloid is $x^2 + y^2 = 3z$
I've already done for the cylindrical coordinates and got:
$$\int^{2\pi}_0\int_0^1\int_{\frac{r^2}{3}}^\sqrt{4-r^2}rdzdrd\theta = \frac{\pi(31-12\sqrt{3})}{6}$$
But I'm having troubles when building the integral for spherical coordinates. So far I've found the bounds of integration for $d\rho$ and $d\theta$.
$$\int^{2\pi}_0\int\int_0^2\rho ^2\sin\phi \,\,\,d\rho d\phi d\theta$$
What about the bounds for $d\phi$? Done a small search and had to solve $3\rho\cos\phi = \rho^2\sin^2\theta$, but this is the point where I'm stuck. Is this right? If yes, how to solve this?
 A: In cylindrical coordinates, the volume is
$$\int^{2\pi}_0 \int_0^{\sqrt3} \int^{-\frac{r^2}{3}}_{-\sqrt{4-r^2}} r\>dr d\theta 
= \frac{19\pi}6$$
and, in spherical coordinates
$$\int^{2\pi}_0 \int_{\pi/2}^{2\pi/3} \int_0^{-\frac{3\cos\phi}{\sin^2\phi}}
\rho ^2\sin\phi \,d\rho d\phi d\theta
+\int^{2\pi}_0 \int^{\pi}_{2\pi/3} \int_0^{2}
\rho ^2\sin\phi \,d\rho d\phi d\theta = \frac{19\pi}6$$
A: At the intersection of paraboloid and sphere,
$x^2 + y^2 + z^2 = 4, x^2+y^2 = 3z \implies z^2 + 3z - 4 = 0$.
i.e, $z = 1, x^2+y^2 = 3$.
So in cylindrical coordinates, $0 \leq r \leq \sqrt3$.
Now in spherical coordinates,
$x = \rho \cos\theta \sin\phi,y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
At the intersection of paraboloid and sphere,
$\rho = 2, z = \rho \cos\phi = 1,  \implies \cos\phi = \frac{1}{2}$
So for $0 \leq \phi \leq \frac{\pi}{3}$, the region is bound by the sphere and for $\frac{\pi}{3} \leq \phi \leq \frac{\pi}{2}$, the region is bound by the paraboloid.
$3z = x^2 + y^2 \implies 3 \rho \cos\phi = \rho^2 \sin^2\phi$
$\rho = 3 \cos\phi \csc^2\phi$
So bounds for the second integral are,
$\frac{\pi}{3} \leq \phi \leq \frac{\pi}{2}, 0 \leq \rho \leq 3 \cos\phi \csc^2\phi $
