3
$\begingroup$

I am trying to calculate the flux through the unit sphere centered at the origin given a vector field $F:\mathbb{R}^3 \setminus \{(0,0,0)\} \rightarrow \mathbb{R}^3$ with $\operatorname{div} F=1/(x^2+y^2+z^2)$.

I can't apply the divergence theorem directly because of the discontinuity so what I have done instead is to consider an inner sphere of infinitesimal radius $\epsilon>0$ and write

$\iiint_V \operatorname{div} F dV$= (flux through unit sphere)-(flux through inner sphere).

However, I really don't know how to go on from here. Can I prove that the flux through the sphere of radius $\epsilon$ is $0$ or is this not even true?

I would appreciate some help.

$\endgroup$
  • $\begingroup$ You can start computing the inner flux. Just consider the integral $$\int1/r^2 drd\Omega$$. $\endgroup$ – Dog_69 Sep 3 '18 at 16:33
  • $\begingroup$ @Dog_69 do you mean calculating the flux of the inner sphere of radius $ \epsilon$? Doesn't this calculation require the use of the divergence theorem for the surface of the inner sphere? If yes, I don't think it's possible to do it that way because the origin is included in that sphere too and the vector field is not defined there. $\endgroup$ – DreamCream Sep 3 '18 at 19:54
  • 1
    $\begingroup$ Oh! You're right. I was thinking of computing the integral from $\epsilon$ to $R$ but I wrote something completely different. So that, I would try to compute the above integral (notice that a factor $r^2$ is missed) from $\epsilon$ to $R$ to get $4\pi(R-\epsilon)$ and then take the limit $\epsilon\to 0$. $\endgroup$ – Dog_69 Sep 4 '18 at 8:03
2
$\begingroup$

For points ${\bf x}=(x_1,x_2,x_3)\in{\mathbb R}^3$ write $|{\bf x}|=:r$, and consider the vector field $${\bf v}({\bf x}):=\left({c\over r^2}+{1\over r}\right)\>{{\bf x}\over r}\qquad({\bf x}\ne{\bf 0})\ ,$$ where $c$ is an arbitrary constant. One computes$${\partial v_1\over\partial x_1}=-\left({3c\over r^4}+{2\over r^3}\right){x_1\over r}x_1+\left({c\over r^3}+{1\over r^2}\right)\ ,$$ which then leads to $${\rm div}\,{\bf v}({\bf x})={1\over r^2}\qquad({\bf x}\ne{\bf 0})\>,$$ as required. Note that the part ${\displaystyle{c\over r^2}{{\bf x}\over r}}$ of ${\bf v}$, the gravitational field of a mass point at ${\bf 0}$, has zero divergence. The field ${\bf v}$ is radial and with constant radial component on spheres $S_R$ centered at ${\bf 0}$. The flux $\Phi$ of ${\bf v}$ through such a sphere therefore is given by $$\Phi(S_R)=\left({c\over R^2}+{1\over R}\right)\>{\rm area}(S_R)=4\pi(c+R)\ .$$It follows that this flux is not determined by the data given in the problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.