Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 3 / 11B / Solution

Using the Levi-Civita symbol and the epsilon-delta identity,ThereforeOne particularly simple vector potential for $u$ isDirect differentiation gives $\nabla \cdot A=0$ and $\nabla \times A=(y^2,z^2,x^2)=u$.

Parametrize the curved cone byThe outward oriented surface element isSubstitution and integration over $0\le z\le h$, $0\le \phi <2\pi$ give

Because $\nabla \cdot u=0$, the divergence theorem predicts that the two outward fluxes sum to zero. On the top disk $S_2$, $d\mathbf{S}=(0,0,1)dxdy$ and $u_z=x^2$, so, with $R=h\tan \alpha$,as predicted.

The outward orientation on $S_2$ induces the stated anticlockwise orientation on $C$, whereas the outward orientation on $S_1$ induces the reverse orientation. Since $u=\nabla \times A$, Stokes theorem predictsOn $C$, put $(x,y,z)=(R\cos \phi ,R\sin \phi ,h)$. Thenverifying both identities.