Given the potential function:
$$\Phi(a,b,c) = \iiint\limits_{V} \frac {1}{\sqrt {(x-a)^2 + (y-b)^2+(z-c)^2}}dx\ dy\ dz$$
for every $ (a,b,c) \notin V $ and $V = \{(x,y,z)|\space R_1 \leq x^2+y^2+z^2 \leq R_2 \}$
I need to calculate $\Phi(0,0,c)$ for $0 < c < R_1 $ and for $R_2 < c$.
I tried using spherical coordinates, but it only helps calculating $\Phi(0,0,0)$, which is $2\pi(R_2^2-R_1^2)$
if I'm not wrong, since the integral at $(0,0,c)$ becomes $$ \int_0^{2\pi}d\theta \int_0^{\pi}d\phi \int_{R_1}^{R_2} \frac {r^2 \cdot \sin\phi}{\sqrt{r^2-2c\cdot \cos\phi \cdot r + c^2}}dr$$ and I don't know how to solve it.
also, the result should be that the potential within all of the sphere $B((0,0,0),R_1)$ is constant.
I guess there's a simpler way to do it which I am missing (maybe another type of variable substitution? or using the second type of line integral $ \int\limits_{(0,0,0)}^{(0,0,c)}\nabla\phi \cdot dr $ ? ), so any idea will be helpful.
integrate with respect to $\phi$ first:
$\iiint \frac {r^2\sin\phi}{\sqrt{r^2-2cr\cos\phi + c^2}} \ d\phi\ dr\ d\theta\\ \iint \frac {r\sqrt{r^2-2cr\cos\phi + c^2}}{c}|_0^\pi \ dr\ d\theta\\ \iint \frac {r (|r+c| - |r-c|)}{c} \ dr\ d\theta$
If $c < R_1$ then $|r+c| - |r-c| = 2c$ over the interval $[R_1,R_2]$
if $c >R_2$ then $|r+c| - |r-c| = 2r$ over the inteval.