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. 
 A: 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 $\phi \le \frac {\pi}{2}$
$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{4\cos\phi} (\rho^3)(\rho\sin\phi) \ d\rho\ d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} (\frac 15) (4^5) (\cos^5 \phi)(\sin\phi) d\phi\ d\theta\\
\int_0^{2\pi}\int_0^{\frac {\pi}{2}} -(\frac 1{30}) (4^5) \cos^6 \phi\ d\theta\\
(2\pi)(\frac {1}{30})4^5$
A: The first thing to notice here is exactly what volume you're integrating inside.  The volume you're integrating is the sphere of radius $2$ but centered at the point $(0,0,2)$.  If we converted directly to spherical coordinates for this volume, our limits will indeed be very tricky.  What I would recommend instead is to translate your coordinate system from $(x,y,z)$ to $(x',y',z')$ where
$$
x'=x \hspace{2pc} y'=y \hspace{2pc} z' = z-2.
$$
In this case your differentials all stay the same, but instead what changes is your integrand.  Instead it changes to:
$$
\left (x'^2 + y'^2 + (z'+2)^2  \right )^{3/2} \;\; =\;\; \left (x'^2 + y'^2 + z'^2 + 4z' + 4 \right )^{3/2}.
$$
With this transformation we find that the limits of integration will transform to
$$
\int_{-2}^2 \int_{-\sqrt{4-x’^2}}^{\sqrt{4-x’^2}} \int_{-\sqrt{4-x’^2-y’^2}}^{\sqrt{4-x’^2-y’^2}} \left (x'^2 + y'^2 + z'^2 + 4z' + 4 \right )^{3/2} dz'dy'dx'.
$$
At least here you get to integrate over a sphere centered at the origin.  Your integrand will transform as
$$
\left (x'^2 + y'^2 + z'^2 + 4z' + 4 \right )^{3/2} dz'dy'dx' \;\; \to \;\; \left (\rho^2 + 4\rho\cos\theta + 4\right )^{3/2}\rho^2\sin\theta d\rho d\theta d\phi.
$$
How to integrate this expression is an added difficulty, but at least the bounds are reasonable.
