Let $F ( x ) = \frac { x } { \| x \| ^ { 3 } }$, if $0 \in \Omega$ show $\iint _ { \partial \Omega } ( F \cdot \nu ) d s = 4 \pi$ Problem: Let $\Omega \subset \mathbb { R } ^ { 3 }$ a regular domain and $\nu$ the unit external vector of $\partial \Omega$. Let $F ( x ) = \frac { x } { \| x \| ^ { 3 } }$. 
Prove that if $0 \in \Omega$, then $\iint _ { \partial \Omega } ( F \cdot \nu ) d s = 4 \pi$.
I tried to apply the divergence theorem, $$\iiint _ { \Omega } \operatorname { div } F \mathrm { d } x _ { 1 } \mathrm { d } x _ { 2 } \mathrm { d } x _ { 3 } = \iint _ { \partial \Omega } ( F \cdot \nu ) \mathrm { d } s$$
But without getting interesting results since I have so few informations about $\Omega$.
 A: You cannot apply divergence theorem directly to the domain $\Omega$ since $F$ is not $C^1$ on the whole domain $\Omega$ due to $0\in\Omega$. Instead, let $B_\epsilon =\{x\in\mathbb{R}^n\;|\;\|x\|<\epsilon\}$. Pick sufficiently small $\epsilon>0$ such that $B_{2\epsilon}\subset \Omega$ and let $\Omega_\epsilon = \Omega \setminus B_\epsilon$. Then, $\partial\Omega_\epsilon= \partial \Omega -\partial B_\epsilon$ and it holds
$$
\int_{\Omega_\epsilon}\text{div}F\;dxdy=\int_{\partial\Omega_\epsilon}F\cdot\nu \;dS =\int_{\partial\Omega}F\cdot\nu \;dS-\int_{\partial B_\epsilon}F\cdot\nu \;dS.
$$ Note that $\text{div}F=0$ on $\Omega_\epsilon$ and hence 
$$
\int_{\partial\Omega}F\cdot\nu \;dS=\int_{\partial B_\epsilon}F\cdot\nu \;dS.
$$ Now, since outward unit normal vector $\nu(x)$ on $\partial B_\epsilon$ is $\frac{x}{\epsilon}$, we get
$$
\int_{\partial B_\epsilon}F\cdot\nu \;dS=\int_{\partial B_\epsilon}\frac{x}{r^3}\cdot\frac{x}{\epsilon} \;dS=\frac{1}{\epsilon^2}\text{vol}(\partial B_\epsilon)=4\pi.
$$
