Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 2 / a / iv

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.