Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 19

For a smooth oriented surface $S$ with positively oriented boundary $C=\partial S$, Stokes theorem statesHere $d\mathbf{S}=\mathbf{n}dA$ uses the chosen unit normal vector, while $d\mathbf{r}$ is tangent to the directed boundary. The boundary orientation follows the right-hand rule: curling the fingers along $C$ makes the thumb point in the normal direction. The field must be continuously differentiable on a neighbourhood of the surface, with the usual piecewise-smooth allowance for its boundary.

For the upper hemisphere with outward normal, the induced equatorial direction is counterclockwise as viewed from positive $z$.

First calculate the curl:By Stokes theorem, use the positively oriented equator $\mathbf{r}(\varphi )=(a\cos \varphi ,a\sin \varphi ,0)$, $0\le \varphi \le 2\pi$. Its tangent is $a(-\sin \varphi ,\cos \varphi ,0)$, while the restricted field is $(0,3a^2\cos \varphi \sin \varphi -a\cos \varphi ,0)$. ThusThe first term integrates to zero over a full period and the second uses ${\int }_0^{2\pi }{\cos }^2\varphi d\varphi =\pi$.

One can also calculate directly on the curved surface. With spherical coordinates $0\le \theta \le \pi /2$, $0\le \varphi \le 2\pi$,Substitution gives the scalar integrand $a^3{\sin }^2\theta \cos \theta \sin \varphi -a^2\sin \theta \cos \theta$. Its azimuthal sine term vanishes, and the rest gives $-2\pi a^2{\int }_0^{\pi /2}\sin \theta \cos \theta d\theta =-\pi a^2$, confirming the orientation and the same result.