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

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.