Let $S$ a sphere or radius $r$ centered at the origin and $\vec{p}$ which is inside or outside but not in the sphere. Show that
$$\int \int _S \frac{1}{||\vec{x}-\vec{p}||}dS=\begin{cases}
4\pi r \hspace{10 pt} \text{if ||p||<r },\\
\frac{4\pi r^2}{||p||} \hspace{10 pt} \text{if ||p||>r }
\end{cases}$$
I tried to show it by the parametrization of $S$, and using that $||\vec{x}-\vec{p}||=\sqrt{||\vec{p}||^2+r^2-2||\vec{p}||r \langle p,\vec{x}\rangle}$, I used mean value too but without results (I got $\frac{4\pi r^2 }{||\vec{{{q}}}-\vec{p}||}$, $||q||=r$ ). And geometrically I can't see how $\frac{1}{||\vec{x}-\vec{p}||}$ changes when $\vec{p}$ is inside or outside. Any help is appreciate
Best Answer
As mentioned in a comment above, this question has been answered here.
The trick is to do the integral in spherical coordinates, and to simplify the calculations with symmetry arguments: