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 ?

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{}$
\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}
