Prove $\iiint_V\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dV$ is constant inside a ball I have been struggling with this problem for a while:
Let $V$ be the volume:
$$V=\{(x,y,z)| R_1^2\leq x^2+y^2+z^2\leq R_2^2\}$$
Such that $0<R_1 <R_2$.
We will define a new function $\phi(a,b,c)$, which is defined for every $(a,b,c)\notin V$:
$$\phi(a,b,c)=\iiint_V\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dxdydz$$
Our task is to prove that $\phi(a,b,c)$ is constant  inside the ball $B((0,0,0),R_1)$.
I tried to change the variables using spherical coordinates:
$$x=r\cos\varphi\sin\theta$$
$$y=r\sin\varphi\sin\theta$$
$$z=r\cos\theta$$
$$r\in[R_1,R_2], \theta \in[0,\pi] ,\varphi \in[0,2\pi]$$
And then solve the integral, proving it is constant when $r<R_1$, but the integral was a bit hard to solve. I assume there's an easier way - but I couldn't think of one.
Thanks!
P.S. - Yes, I noticed the Physics here - electric potential inside a ball! But unforunately this is not the course - I have to be rigorous.
 A: Solution 1.
Let $\ell = \sqrt{a^2 + b^2 + c^2}$. By a suitable rotation, it suffices to assume that $(a, b, c) = (0, 0, \ell)$. Then applying the spherical coordinates,
\begin{align*}
\phi(a, b, c)
&= \int_{R_1}^{R_2} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \int_{0}^{2\pi} \mathrm{d}\varphi \, \frac{r^2 \sin\theta}{\sqrt{r^2 - 2r\ell \cos\theta + \ell^2}} \\
&= \int_{R_1}^{R_2} \mathrm{d}r \int_{0}^{\pi} \mathrm{d}\theta \, \frac{2\pi r^2 \sin\theta}{\sqrt{r^2 - 2r\ell \cos\theta + \ell^2}} \\
&= \int_{R_1}^{R_2} \mathrm{d}r \, \left[ \frac{2\pi r}{\ell} \sqrt{r^2 - 2r\ell \cos\theta + \ell^2} \right]_{\theta = 0}^{\theta = \pi} \\
&= \int_{R_1}^{R_2} \mathrm{d}r \, \frac{2\pi r}{\ell} (r + \ell - |r - \ell|).
\end{align*}
Since $\ell < R_1 \leq r$, it follows that $|r - \ell| = r - \ell$, and so,
\begin{align*}
\phi(a, b, c)
= \int_{R_1}^{R_2} \mathrm{d}r \, 4\pi r
= 2\pi (R_2^2 - R_1^2).
\end{align*}
This is independent of $\ell$.

Solution 2.
Notice that $\phi$ is harmonic on $B(0, R_1)$ and continuous on $\overline{B(0, R_1)}$. By the maximum principle, both the maximum and the minimum of $\phi$ are achieved on the boundary $\partial B(0, R_1)$. But by the rotational symmetry, $\phi$ is constant on $\partial B(0, R_1)$. Therefore $\phi$ is constant on all of $\overline{B(0, R_1)}$.
