# 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.

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So you can compute this thing on the sphere directly without resorting to Divergence theorem. For any surface, think of it as the outer boundary of a solid. I am with holding one more hint unless you really need it. –  Braindead Mar 23 '13 at 16:46

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 $$\int\nolimits_S {\bf v}\cdot\vec{d\omega}=\int\nolimits_{\partial B_\epsilon} {\bf v}\cdot\vec{d\omega}=\int\nolimits_{\partial B_\epsilon} {1\over\epsilon^2}\ {\rm d}\omega=\ \omega(\partial B_\epsilon)\ {1\over\epsilon^2}=4\pi\ .$$
Why do you have $\mathrm{div}(v) = 0$? In order to know this, you have to extend $\hat \rho$ to $B$. –  gerw Mar 23 '13 at 17:33
@gerw No you don't. The integral is taken over $B'$, and the Gauss theorem is being applied to the integral taken over $B'$, not $B$. –  Braindead Mar 23 '13 at 18:30
Ok, this was a typo. You have to extend $\hat\rho$ to B'. –  gerw Mar 23 '13 at 18:40
I think I've understood what he did. $B'$ doesn't contain the origin, so in $B'$ the divergence is zero and the theorem really holds. That's the key to the proof right ? We simply "cut off" where there's the problem, and then we use the relations involving the integrals over the sets to show what's desired. Thanks for the answer! –  user1620696 Mar 23 '13 at 19:00
@ChristianBlatter: Now I've realized what's going on here. By $\hat\rho$ you didn't denote the outer normal vector (of the surface) as written by the OP, but the unit vector pointing away from the origin. You are totally right! Sorry for the inconvinience. –  gerw Mar 24 '13 at 17:44