Triple Integrals in Spherical Coordinates where (z-2)^2 $$\iiint \frac{1}{\sqrt{x^2+y^2+(z-2)^2}}$$
for $x^2+y^2+z^2 = 1$
I've used spherical coordinates, like this:
$x=\rho sin\phi cos\theta$; $y=\rho sin\phi sin\theta$; $z=\rho cos\phi$ and $J=\rho^2 sin\phi$
$$\rho^2 sin^2\phi cos^2\theta + \rho^2 sin^2\phi sin^2\theta +\rho^2 cos^2\phi = 1$$
$$\rho^2 sin^2\phi + \rho^2 cos^2\phi = 1$$
$$\rho^2 = 1$$
$$\rho = 1$$
and when i put coordinates in my initial integral and simplify it, i get this:
$$\int_0^{2\pi}\int_0^\pi\int_0^1\frac{1}{\sqrt{\rho^2-4\rho cos\phi +4}}\rho^2sin\phi d\theta d\phi d\rho$$
and i can not move from there. 
Please, help.
 A: $$\begin{align}&\int_0^{2\pi}\int_0^{\pi}\int_0^1\frac1{\sqrt{\rho^2-4\rho\cos\phi+4}}\rho^2d\rho\sin\phi\,d\phi\,d\theta\\
&=2\pi\int_0^{\pi}\int_0^1\frac1{\sqrt{\rho^2-4\rho\cos\phi+4}}\rho^2d\rho\sin\phi\,d\phi\\
&=2\pi\int_0^1\left[\frac12\sqrt{\rho^2-4\rho\cos\phi+4}\right]_0^{\pi}\rho\,d\rho\\
&=\pi\int_0^1\left[|\rho+2|-|\rho-2|\right]\rho\,d\rho\\
&=\pi\int_0^12\rho^2d\rho=\frac{2\pi}3\end{align}$$
A: I was going to suggest the substitution.
$\rho = 2\sin \phi\tan u + 2 \cos \phi$
And that would clean up the radical.  But, after thinking about it, I decided that spherical is not the way to go.
Suppose instead you did cylindrical.
$x = r \cos \theta\\
y = r \sin \theta\\
z = z\\
dx\,dy\,dz = r\,dr\,dz\,d\theta$
$\int\int\int\dfrac {r}{\sqrt{r^2 + (z-2)^2}} dr\,dz\,d\theta$
We would like to integrate by r first.  Otherwise, we have an inverse tangent to deal with.  So, find limits of r in terms of z.
$\int_0^{2\pi}\int_{-1}^1\int_0^{\sqrt{1-z^2}} \dfrac {r}{\sqrt{r^2 + (z-2)^2}} dr\,dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1  (r^2 + (z-2)^2)^{\frac12}|_0^{\sqrt{1-z^2}} dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1  [(1-z^2 + (z-2)^2)^{\frac12} - ((z-2)^2)^{\frac12}]dz\,d\theta\\
\int_0^{2\pi}\int_{-1}^1  (-4z+5)^{\frac12} - |z-2|dz\,d\theta$
And you can get home from here.
Suppose we stuck with spherical, and flipped the order of integration!
$\int_0^1\int_0^\pi\int_0^{2\pi}\frac{\rho^2\sin\phi}{\sqrt{\rho^2-4\rho \cos\phi +4}} d\theta d\phi d\rho$
$2\pi\int_0^1\int_0^\pi\frac{\rho^2sin\phi}{\sqrt{\rho^2-4\rho \cos\phi +4}} d\phi d\rho$
$u = \rho^2-4\rho \cos\phi +4\\
du = 4\rho \sin\phi d\phi$
$2\pi\int_0^1\int_{\rho^2 - 4\rho + 4}^{\rho^2+4\rho + 4}\frac{\rho}{4\sqrt{u}} du d\rho\\
2\pi\int_0^1\frac 12 \sqrt{u}\rho|_{\rho^2-4\rho + 4}^{\rho^2+4\rho + 4} d\rho\\
2\pi\int_0^1\frac 12 (|\rho+2|\rho - |\rho-2|\rho) d\rho\\
2\pi\int_0^1 \rho^2\, d\rho\\
\frac{2\pi}{3} \rho^3|_0^1 \\
\frac{2\pi}{3}$
A: I think it would be better if you first use $t=z-2$. This way, you have $(x^2+y^2+z^2=1)\rightarrow (x^2+y^2+t^2+4t+3=0)$. Then, use the spherical coordinates. Now, your integral would be really simple, and you need to find the proper boundaries for new integral. 
$(x^2+y^2+t^2+4t+3=0) \rightarrow \rho^2+4\rho cos\phi +3=0 \rightarrow \rho =-2cos\phi \pm \sqrt{4(cos\phi)^2-3} \rightarrow -2cos\phi - \sqrt{4(cos\phi)^2-3}\leq \rho\leq  -2cos\phi + \sqrt{4(cos\phi)^2-3}$.
