Volume enclosed by a paraboloid, cylinder and a plane by Spherical coordinates I am stuck over this triple integral where i have to find  following:
$\int\int\int_V \space z^2 \space dxdydz \space$ over a volume bounded by cylinder $x^2 + y^2 = a^2$, a paraboloid $x^2 + y^2 = z$ and a plane $z=0$.
I did it from cylindrical coordinates. However i want to do it with spherical coordinates. Any help is much appreciated.
 A: The bounds convert to spherical like this:
$$\begin{cases}x^2+y^2 = a^2 \\ x^2+y^2 = z \\ z = 0 \\ \end{cases} \implies \begin{cases}\rho\sin\phi = a \\ \rho\sin^2\phi = \cos\phi \\ \phi = \frac{\pi}{2} \\  \end{cases}$$
We could set up the integral now with $\rho$ first
$$I = \int_0^{2\pi} \int_{\cot^{-1}(a)}^{\frac{\pi}{2}} \int_{\frac{\cos\phi}{\sin^2\phi}}^{\frac{a}{\sin\phi}} \rho^4 \cos^2\phi\sin\phi\:d\rho\:d\phi\:d\theta$$
but this is a mess of trig. The intuition here is when you see lots of trig in the bounds for spherical, try to do $\phi$ first, even though that means we will have to do two integrals
$$I = \int_0^{2\pi}\int_0^a \int_{\cos^{-1}\left(\frac{-1+\sqrt{1+4\rho^2}}{2\rho}\right)}^{\frac{\pi}{2}} \rho^4 \cos^2\phi\sin\phi\:d\rho\:d\phi\:d\theta + \int_0^{2\pi}\int_a^{a\sqrt{a^2+1}}\int_{\cos^{-1}\left(\frac{-1+\sqrt{1+4\rho^2}}{2\rho}\right)}^{\sin^{-1}\left(\frac{a}{\rho}\right)}\rho^4\cos^2\phi\sin\phi\:d\rho\:d\phi\:d\theta$$
$$= \frac{\pi}{12}\left[\int_0^a\rho\left(-1+\sqrt{1+4\rho^2}\right)^3\:d\rho + \int_a^{a\sqrt{a^2+1}}\rho\left(-1+\sqrt{1+4\rho^2}\right)^3-8\rho\left(\rho^2-a^2\right)^{\frac{3}{2}}\:d\rho\right]$$
I will let you take it from here.
A: Hints: Spherical Coordinates
$$x=r sin \theta \cos \varphi, \space y =r \sin \theta \sin \varphi, \space z=r \cos \theta, \space J=r^2 \sin \theta.$$
$V: \space 0 \le r \le \frac{\cos \theta}{\sin^2 \theta}, \space 0 \le \varphi \le 2\pi, \space 0 \le \theta \le \frac{\pi}{2}.$
Figure:

