Application of the divergence theorem

Question

Suppose that $\Omega$ is a bounded domain in $\mathbb{R}^3$ with $C^1$ boundary. I want to prove that $$\int_{\partial \Omega} \nu(y) \cdot \frac{y}{|y|^3} \, dS(y)=\left\{\begin{array}{cc}0 & 0 \in \mathbb{R}^3 \setminus \overline{\Omega} \\ 4\pi & 0 \in \Omega \end{array}\right.$$ where $\nu(y)$ is the unit exterior normal, and $dS$ is surface measure on $\partial\Omega$. If $0 \in \mathbb{R}^3\setminus\overline{\Omega}$ we'll use the divergence theorem applied to the $C^1$ vector field $F(y)=y/|y|^3$. In the case $0 \in \Omega$, we'll do the same thing to the vector field $y/|y|^3 \in C^1(\Omega\setminus\overline{B(0,\varepsilon)})$ for $\varepsilon>0$ sufficiently small. In either case, the result should follow from $\operatorname{div} F=0$, but I am unable to show this. Can someone help?

Answer 1

First notice that $y/|y|^3=\nabla (|y|^{-1})$ so that your integrand is $\partial_\nu f$, with $f(y)=1/|y|$.

If $0\in \mathbb{R}^3\setminus \bar{\Omega}$, then $f$ is harmonic in $\Omega$ so that the divergence theorem gives $$ \int_{\partial \Omega} \partial_\nu f(y) dS = \int_\Omega \Delta f =0. $$ If $0\in \Omega$ and $\Omega_\varepsilon=\Omega\setminus B(0,\varepsilon)$, then $f$ is harmonic in $\Omega_\varepsilon$ so that by the divergence theorem $$ 0= \int_\Omega \Delta f = \int_{\partial \Omega_\varepsilon} \partial_\nu f(y) dS $$ The last integral is equal to $$ \int_{\partial \Omega} \partial_\nu f(y) dS - \int_{\partial B(0,\varepsilon)} \partial_\nu f(y) dy. $$ Now the integral over the boundary of the ball we can compute since $\nu(y)=y/\varepsilon$ and we get $$ \int_{\partial B(0,\varepsilon)} \partial_\nu f(y) dy= \int_{\partial B(0,\varepsilon)}\frac{1}{\varepsilon^2}= \frac{4\pi \varepsilon^2}{\varepsilon^2}=4\pi. $$ Combining all this, we get the result.