Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 1 / c

Let $a=\sinh 1$ and $c=\cosh 1$. The region is the first-octant portion of the ellipsoid $(x^2+y^2)/a^2+z^2/c^2\le 1$. In cylindrical coordinates, write $\rho =\sqrt{x^2+y^2}$ and $0\le \varphi \le \pi /2$. Away from the axis, the divergence isThe field is singular on the axis, so applying the classical divergence theorem there without qualification would be invalid. Instead remove $\rho <\epsilon$ and apply it to the remaining region. On the added cylindrical boundary the unit normal is $-{\mathbf{e}}_{\rho }$, andIts area element is $\epsilon d\varphi dz$, so its flux is $O(\epsilon |\log \epsilon |)\to 0$. The axial component of the field has zero dot product with this normal. The volume integral of the divergence is locally integrable, because its $1/\rho$ factor cancels the cylindrical measure. Thus removing a singular axis in a flux integral gives the desired improper surface integral with no extra axial term.

The divergence theorem now reduces the calculation toThe integral over $s$ is the area of a quarter unit disk. All boundary fluxes are interpreted as convergent improper integrals at the axis.