Proving that the value of the integral doesn't depend on the surface I'm trying to prove that if we have the vector field $v : \mathbb{R}^n \to \mathbb{R}^n$ given in spherical coordinates by:
$$v(\rho, \theta, \phi)=\frac{1}{\rho^2}\hat{\rho}$$
Where $\hat{\rho}$ is the outward unit vector, then the integral of this field over any surface should be $4\pi$. I don't know if it's really true, but it seems true for me. I'm trying to prove or give a counterexample, but I didn't find a proof neither a counterexample.
Can someone give a hint on how to get started with this problem ? Thanks very much for your help!
Edit: I'm thinking on a closed surface enclosing the origin. I know that the field is not defined there, and that is the reason I'm trying to prove that independent of the closed surface that encloses the origin the integral should be $4\pi$. Sorry if I forgot to mention this before.
 A: Let $S$ be a "closed surface enclosing the origin" $O$, oriented outwards. Such an $S$ bounds a certain "three-dimensional body" $B$ which contains  $O$. This idea includes, e.g. the boundary $\partial C$ of the cube $C:=[-1,1]^3$. Then there will be an $\epsilon>0$ such that all points of $S$ 
have a distance $>\epsilon$ from $O$. Denote by $B_\epsilon$ the open ball with radius $\epsilon$ and center $O$. 
Now we apply Gauss' theorem to the body $B':=B\setminus B_\epsilon$. The set $B'$, whose  boundary is given by $\partial B'=S-\partial B_\epsilon\ $, no longer contains $O$. Therefore it is allowed to apply Gauss' theorem to $B'$ and ${\bf v}$. We obtain
$$\int\nolimits_S {\bf v}\cdot\vec{d\omega}-\int\nolimits_{\partial B_\epsilon} {\bf v}\cdot\vec{d\omega}=\int\nolimits_{\partial B'} {\bf v}\cdot\vec{d\omega}=\int\nolimits_{B'}{\rm div}({\bf v})\ {\rm dvol}=0\ ,$$
which implies
$$\int\nolimits_S {\bf v}\cdot\vec{d\omega}=\int\nolimits_{\partial B_\epsilon} {\bf v}\cdot\vec{d\omega}=\int\nolimits_{\partial B_\epsilon} {1\over\epsilon^2}\ {\rm d}\omega=\ \omega(\partial B_\epsilon)\ {1\over\epsilon^2}=4\pi\ .$$
