Past exam of the mathematical methods II course of the University of Oxford / 2017 / 8

Let $a=h/2$ and place the hole axis along $z$. At the rim, $r_0^2=R^2-a^2$. A horizontal slice at $|z|\le a$ is an annulus of area $\pi [(R^2-z^2)-r_0^2]=\pi (a^2-z^2)$. Thus the napkin ring problem givesEach slice depends only on $a$ and $z$, explaining the independence from the original sphere radius, rather than obtaining it merely by cancellation after integration.

The divergence theorem states ${\iint }_{S}F\cdot ndS={\iiint }_V\nabla \cdot FdV$ for a sufficiently smooth field and a piecewise smooth closed boundary with outward unit normal. Choose $F=r/3$. Since $\nabla \cdot r=3$,The factor $1/3$ comes from the three spatial dimensions; the outward orientation is especially important on the inner wall of the drilled solid.

The boundary consists of the spherical zone $-a\le z\le a$ and the inner cylindrical wall. On the sphere, $n=r/R$, so $n\cdot r=R$. Its area element is $Rd\varphi dz$, giving the fluxOn the inner cylinder, the outward normal of the remaining material points towards the hole: $n=-e_{\rho }$. Therefore $n\cdot r=-r_0$, and $dS=r_{0}d\varphi dz$, giving flux $-4\pi a{r}_0^2$. There are no planar end faces of positive area: the two surfaces meet along the rim circles. HenceUsing the cylinder's radial normal with the opposite sign would count the hole as extra material and lose the required cancellation.