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

The three coordinate basis vectors are the partial derivatives of the position vector with respect to the three coordinates. Direct differentiation givesThe multivariable chain rule then gives $d\mathbf{x}={\mathbf{h}}_{\xi }d\xi +{\mathbf{h}}_{\theta }d\theta +{\mathbf{h}}_{\varphi }d\varphi$. These are coordinate tangent vectors, not necessarily unit vectors.

The azimuthal vector ${\mathbf{h}}_{\varphi }$ is automatically orthogonal to the other two. Their remaining dot product isThus orthogonal coordinates require $g{g}^{\prime }=\sinh \xi \cosh \xi$, which integrates to $g^2={\sinh }^2\xi +C$. The initial value gives $C=1$, and the smooth branch with $g(0)=1$ isThe corresponding coordinate-vector norms are $|{\mathbf{h}}_{\xi }|=|{\mathbf{h}}_{\theta }|=\sqrt{{\sinh }^2\xi +{\sin }^2\theta }$ and $|{\mathbf{h}}_{\varphi }|=\sinh \xi \sin \theta$, as in the hyperbolic parametrization of prolate spheroidal coordinates. Coordinate degeneracies on the axis and focal segment are understood.

Eliminating the angular coordinates on the requested surface givesIt is a prolate spheroid with long axis $z$, equatorial radius $\sinh 1$, polar semiaxis $\cosh 1$, and foci $(0,0,\pm 1)$.

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.