Vector field ${\bf F}$ with $\int_S {\bf F}\cdot{\bf n}\ dS=c$ Find a vector 
field ${\bf F}$ on $
{\bf R}^3$ with $$\int_S {\bf F}\cdot{\bf n}\ dS=c > 0 \tag{1} $$ where $S$ is any closed surface containing $0$ and ${\bf n}$ is 
normal 
Here there is a solution $\frac{k}{r^3} (x,y,z)$. Note that from divergence therem we know that ${\bf F}$ has
divergence $0$ so that $(x,y,-2z)$ is possible. But this solution does not satisfy $(1)$. 
So the solution is unique ?
 A: Take this one: $\iint_Sr\cdot ndS$ where S is a closed surface as you wish. Indeed, we then have $$\iint_Sr\cdot ndS=\iiint_V\nabla\cdot r dV=\iiint_V\left(\partial_x\text{i}+\partial_y\text{j}+\partial_z\text{k}\right)\cdot(x\text{i}+y\text{j}+z\text{k})dV=3V$$ where $V$ is the volume enclosed by $S$.
A: It is not unique, in fact there are infinitely many possible solutions for any choice of $c>0$. Simply take a vector field $\mathbf{F}$ satisfying:
$$\int_S\mathbf{F}\cdot\mathbf{n}dS=k\neq0$$
and then define your new vector field by $\mathbf{G}=\frac{1}{k}\mathbf{F}$.
Now there are infinitely many such $\mathbf{F}$, since by Stokes theorem you we that:
$$\int_S\mathbf{F}\cdot\mathbf{n}dS=\int_V\operatorname{div}\mathbf{F}dV$$
where $V$ is the volume contained in $S$, and it's easy to find vector fields with e.g. always positive divergence, such as $\mathbf{F}(x,y,z)=(x,0,0)$.
Also, the vector field you gave in your answer is not defined at the origin.
A: When they talk about  "closed surfaces $S$ containing $0$" they tacitly mean that such $S$ should bound a compact body $B\subset{\mathbb R}^3$ which contains $0$ in its interior. Now we cannot have arbitrarily tiny such surfaces giving a fixed value $c>0$ for the integral in question unless something terrible happens at $0$. 
You have remarked that the flow field
$${\bf G}(x,y,z):=\left({x\over r^3},{y\over r^3},{z\over r^3}\right)$$
could play a rôle in this question, and you are right: This field is divergence free outside of $0$ and has flow integral
$$\int_{S_R}{\bf G}\cdot {\bf n}\>{\rm d}\omega=4\pi$$
when integrated over the sphere $S_R$ of radius $R$ centered at $0$. Using Gauss' theorem, applied to $B\setminus S_R$ with $R\ll1$ it follows that the field
$${\bf F}:={c\over4\pi}{\bf G}\tag{1}$$
has flow $c$ through any surface of the kind described in the first paragraph, whence is a solution of the problem.
But this ${\bf F}$ is not the only solution of the problem at hand: Add to ${\bf F}$ an arbitrary $C^1$-vector field ${\bf v}$ which has divergence $0$ in all of ${\mathbb R}^3$. Then ${\bf F}+{\bf v}$ again solves the problem; and it is not difficult to show that all solutions are obtained in this way.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}
&\color{#66f}{\large\int_{V}\nabla\cdot\pars{\vec{r} \over r^{3}}\,\dd V}
=-\int_{V}\nabla\cdot\nabla\pars{1 \over r}\,\dd V
=-\int_{V}\ \overbrace{\ \nabla^{2}\pars{1 \over r}\ }
^{\color{#c00000}{\ds{-\,4\pi\,\delta\pars{\vec{r}}}}}\,\dd V\ =\
4\pi\int_{V}\delta\pars{\vec{r}}\,\dd V
\\[3mm]&=\color{#66f}{\large\left\lbrace\begin{array}{lcl}
4\pi & \mbox{if} & \vec{0}\ \in\ V
\\ 
0 & \mbox{if} & \vec{0}\ \not\in\ V
\end{array}\right.}
\end{align}
