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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.

Amit Zach
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1 Answers1

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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 $\overline{B(0, R_1)}$.

Sangchul Lee
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