Proof using divergence theorem It is given vector field $a=\pm  \frac{x}{| | x |  |^{3} } $
$F\subset \mathbb R ^{3}$ is smooth closed orientable surface with limited range $B\subset \mathbb R ^{3}$ i.e. $F=\partial B$


*

*Show that $diva=0$

*Show following using divergence theorem.


a)$\int_{F}^{} a \!  \, dn=0, 0\notin B$
b)$\int_{F}^{} a \!  \, dn=\pm 4\pi, 0\in B$
I showed that $diva=0$. Can someone help me with $2.$ part of this example?
I know that: $\int_{B}^{} \! diva \, dx=\int_{\partial B}^{} \! a \, dn  $, but how to use it here?
 A: For 2a: 
If $\underline{0} \notin B$ then you can apply the divergence theorem, because the vector field $\underline{a}$ is continuously differentiable in $B$. The theorem says:
\begin{align} \iiint_B \text{div}(\underline{a})\mathrm d V = \iint_{\partial B} \underline{a}\bullet \mathrm d \underline{n}  \end{align}
You have calculated the divergence of $\underline{a}$ and that was equal to zero, so we get:
\begin{align}  \iint_{F} \underline{a}\bullet \mathrm d \underline{n}=\iint_{\partial B} \underline{a}\bullet \mathrm d \underline{n} = \iiint_B 0\mathrm  d V  =0  \end{align}
For 2b:
If $\underline{0} \in B$ then we can not apply the divergence theorem, because $\underline{a}$ is not differentiable in $\underline{0}$.
We know that $F$ is the boundary of the set $B$. Furthermore $F$ is a closed surface, so we can make a ball with radius $\epsilon>0$ and center $\underline{0}$ such that the ball is totally containd in the interior of $B$. (Draw some random surfaces $F$ and verify that you can indeed construct such ball)
Now we have the ball, name it $H$. I only consider the case: $\underline {a}=\frac {\underline{x}}{||\underline{x}||^3} $ and where the orientation of $F$ is such that the normal vector is pointing away from origin. 
The ball has surface $\partial H$. Define $K$ to be the area that is enclosed between the surface of the ball $H$ and $F$. Now we have $K$, we can apply the divergence theorem there, because $\underline{0} \notin K$ and note that $\partial K= \partial H \cup F$. So the theorem says:
\begin{align} \iiint_K \text{div}(\underline{a})\mathrm d V = \iint_{\partial H \cup F} \underline{a}\bullet \mathrm d \underline{n}=\iint_{\partial H } \underline{a}\bullet \mathrm d \underline{n}+\iint_{ F} \underline{a}\bullet \mathrm d \underline{n} \end{align}
We knew that div$(\underline{a})=\underline{0}$,  so:
\begin{align}0=\iint_{\partial H } \underline{a}\bullet \mathrm d \underline{n}+\iint_{ F} \underline{a}\bullet \mathrm d \underline{n}  \Longleftrightarrow \iint_{ F} \underline{a}\bullet \mathrm d \underline{n} = - \iint_{\partial H } \underline{a}\bullet \mathrm d \underline{n}\end{align}
We can calculate the RHS there, because we can parametrise the surface of the ball. Let $\underline{r}(\theta,\phi)$ the parametrisation. 
\begin{align} \underline{r}(\theta,\phi) = \left( \epsilon \sin \phi \cos \theta, \epsilon \sin\phi \sin \theta, \epsilon \cos \phi \right), \hspace{5pt} \theta \in [0, 2\pi],  \phi \in [0, \pi]\end{align}
The normal vector in this paramtrisation is:
\begin{align} \frac{\partial \underline {r}}{\partial \phi}\times \frac{\partial \underline {r}}{\partial \theta}=(\epsilon^2\sin^2\phi \cos \theta, \epsilon^2 \sin^2 \phi \sin \theta, \epsilon^2 \cos\phi \sin \phi) \end{align}
Before putting this all in the integral, note that $||\underline{x}||=\epsilon$ if we are on the surface of the ball: 
\begin{align} \iint_{\partial H } \underline{a}\bullet \mathrm d \underline{n}=\int^{2\pi}_0\int^{\pi}_0 \left(a(\epsilon \sin \phi \cos \theta, \epsilon \sin\phi \sin \theta, \epsilon \cos \phi) , \frac{\partial \underline {r}}{\partial \phi}\times \frac{\partial \underline {r}}{\partial \theta} \right) \mathrm d \phi \mathrm d \theta\end{align}
This becomes:
\begin{align} \iint_{\partial H } \underline{a}\bullet \mathrm d \underline{n}&=\int^{2\pi}_0\int^{\pi}_0 \frac{\epsilon^3 \sin^3 \phi \cos ^2\theta +\epsilon^3 \sin^3\phi \sin^2 \theta + \epsilon^3 \cos^2 \phi\sin\phi}{\epsilon^3}  \mathrm d \phi \mathrm d \theta\\
&=2\pi \int^{\pi}_0 \sin\phi \mathrm d \phi = 4\pi\end{align}
After all this we have: 
\begin{align} \iint_{ F} \underline{a}\bullet \mathrm d \underline{n} =-4\pi \end{align}
The same way we can get $+4\pi$, if we take the other version of $\underline{a}$.
Beautiful thing that can be seen here is that the choice of $\epsilon$ does not even matter after all.
