Let $B(a,r)\subset\mathbb{R}^3$ the ball centered in $a\in\mathbb{R}^3$ with radius $r>0$ and $N$ the normal field of the sphere $S(a,r)$. Show that, if $(0,0,0)\notin S(a,r)$, then $$\int_{S(a,r)}\frac{F(x,y,z)}{|F(x,y,z)|^3}NdS =\begin{cases} 4\pi,& \mbox{if }(0,0,0)\in B(a,r) \\ 0,& \mbox{if }(0,0,0)\notin B(a,r)\end{cases}$$ with $F(x,y,x) = (x,y,z)$ and $|F(x,y,z)|=\sqrt{x^2+y^2+z^2}$
If $(0,0,0)\notin B(a,r)$, then $F$ is $C^1$ and I can use the divergence theorem. So $\nabla .G = 0$, with $G=\frac{F(x,y,z)}{|F(x,y,z)|^3}$
I got stuck when $(0,0,0)\in B(a,r)$. Any help?
Take the volume as $V_{\epsilon} = B(a,r) \setminus B(0,\epsilon) $, $\epsilon$ small enough that the boundary of this is $S(a,r) \cup S(0,\epsilon)$. Then $G$ is $C^1$ on $V_{\epsilon}$, so the divergence theorem gives that the integral over one sphere of the boundary is equal to that over the other sphere, so $$ \int_{S(a,r)} G \cdot N \, dS = \int_{S(0,\epsilon)} G \cdot N \, dS. $$ Now, on the sphere of radius $\epsilon$, $N=(x,y,z)/\epsilon$, while $G=(x,y,z)/\epsilon^3$, since $x^2+y^2+z^2=\epsilon^2$. Then $$ \int_{S(0,\epsilon)} G \cdot N \, dS = \int_{S(0,\epsilon)} \frac{\epsilon^2}{\epsilon^4} \, dS = \frac{1}{\epsilon^2} \int_{S(0,\epsilon)} dS = 4\pi, $$ since the area of the sphere is $4\pi \epsilon^2$.