Very tricky triple integral in spherical coordinates Evaluate $$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{2-\sqrt{4-x^2-y^2}}^{2+\sqrt{4-x^2-y^2}}(x^2+y^2+z^2)^{3/2} \; dz \; dy \; dx$$ by converting to spherical coordinates.
We know that $(x^2+y^2+z^2)^{3/2} = (\rho^2)^{3/2} = \rho^3$. The range of $y$ tells us that we have disk of radius $2$. How can we use this find the limits of each integral and ultimately find the solution? Thanks you for the help. 

The boundary of our sphere is $x^2 + y^2 + z^2 = 4z$
Plugging
$x = \rho\cos\theta \sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho \cos\phi$
We get:
$\rho^2 = 4\rho\cos\phi\\
\rho = 4\cos\phi$
And the sphere is above the $xy$ plane, or $\phi \le \frac {\pi}{2}$
$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{4\cos\phi} (\rho^3)(\rho\sin\phi) \ d\rho\ d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} (\frac 15) (4^5) (\cos^5 \phi)(\sin\phi) d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} -(\frac 1{30}) (4^5) \cos^6 \phi\ d\theta\\
(2\pi)(\frac {1}{30})4^5$
