Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 1

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

For an oriented piecewise smooth surface $S$ with boundary curve $C=\partial S$, and a continuously differentiable vector field $\mathbf{F}$ near $S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is positive surface area, and $d\mathbf{r}$ is the directed tangent line element. The positive traversal of $C$ is fixed by the right-hand rule relative to $\mathbf{n}$. Thus the flux of the curl through the surface equals the circulation along its oriented boundary.

Differentiate the position vector, keeping the other coordinates fixed in each derivative:The first two have zero dot product, since their two planar contributions cancel, and both are orthogonal to ${\mathbf{h}}_z$. Hence these are orthogonal coordinates away from coordinate degeneracies. Their scale factors, which are the lengths of these coordinate basis vectors, areFor example $|{\mathbf{h}}_u{|}^2=a^2[{\sinh }^{2}u{\cos }^{2}v+{\cosh }^{2}u{\sin }^{2}v]=a^2({\sinh }^{2}u+{\sin }^{2}v)$. The unit basis vectors are ${\mathbf{h}}_i/h_i$ when $h_i\ne 0$; the given coordinate basis vectors themselves are not normalized. The Jacobian determinant is $a^2({\sinh }^{2}u+{\sin }^{2}v)>0$ except on the focal lines. These are elliptic cylindrical coordinates.

For $u>0$, eliminating $v$ givesThus constant $u$ surfaces are confocal elliptic cylinders parallel to the $z$ axis. The limiting $u=0$ surface is the vertical strip $y=0$, $|x|\le a$.

For generic $v$, eliminating $u$ givesThese are hyperbolic cylinders with the same focal lines. A specified angular value selects the half-branch with $\mathrm{sgn}x=\mathrm{sgn}(\cos v)$ and $\mathrm{sgn}y=\mathrm{sgn}(\sin v)$, because $u\ge 0$. At $v=0,\pi$ the surfaces are the vertical half-planes $y=0$, $x\ge a$ or $x\le -a$; at $v=\pi /2,3\pi /2$ they are $x=0$, $y\ge 0$ or $y\le 0$.

Finally, constant $z$ gives horizontal planes. The three surface families express the orthogonality of elliptic cylindrical coordinates.

The ellipse bounding the surface is $u=1$. With the upward normal, its positively oriented boundary is traversed by increasing $v$ from zero to $2\pi$:The given components are Cartesian components, so their ordinary dot product with this tangent isTherefore Stokes theorem givesA direct check uses $F_x=2y/a$ and $F_y=-2x/a$, hence $(\nabla \times \mathbf{F}{)}_z=-4/a$, and multiplies by the ellipse area $\pi a^2\sinh 1\cosh 1$. The third component cannot affect either calculation because the surface and its boundary lie at fixed $z$. Equivalently one may apply Stokes to the smooth planar field $(2y/a,-2x/a,0)$, avoiding any need to differentiate the elliptic-coordinate expression for the third component at its foci.