Given the vector field on $R^3$∖{0},
F(x)=$(x/r^3,y/r^3,z/r^3)$, where $r=(x^2+y^2+z^2)^{\frac {1}{2}}$,
let R be a simply connected bounded region, with smooth boundary, in the xy plane containing the origin.
Define for ϵ>0,
$h(ϵ)=∬_RF(x−ϵk)⋅kdS(x)$, where the integral is taken over the region R and k is the unit vector perpendicular to the xy plane.
Compute the jump: $lim_{ϵ↓0}(h(ϵ)−h(−ϵ))$
The hint given in the problem is: Use the divergence theorem in a punctured cap.
My work: First off, the final answer is supposed to be $-4\pi$ , so I am trying to work out the details to show it.
I computed the divergence to be zero -- so F is divergence-free.
Then, since the divergence theorem states that:
$$\int \int_R F.ndS = \int\int\int_{R+\epsilon \bigcup R-\epsilon} (\nabla.F)dV $$
then I am guessing that we can deform both the surface R and the volume it encloses, $R+/-\epsilon$, since the right-hand side of the integration yields zero, since F is divergence-free? So perhaps we can deform the volume to a unit-sphere and work from there to compute $h(\epsilon)$ and subsequently the limit $lim_{ϵ↓0}(h(ϵ)−h(−ϵ))$.
There are some issues, however: At the origin, the vector field blows up. So, since F is not continuously differentiable, the divergence theorem does not apply. I know that the hint says to apply the Theorem in a punctured cap, but I am not sure how to do this.
Finally, this is actually part (b) of the problem statement. For part(a), I computed $$\int\int_{|x|=1} F.ndS = 4\pi,$$
using n = 1/r (x,y,z) as my outward unit-normal, and integrating in spherical-polar coordinates.
I am guessing part(b) likely uses part(a). (That's why I am wondering whether I can deform the volume into a unit-sphere.)
Any hints or solutions are welcome.
Thanks,