Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 10A (Vector Calculus)

Stokes theorem is ${\int }_S(\nabla \times F)\cdot ndS={\oint }_{\partial S}F\cdot dx$, with the boundary orientation given by the right-hand rule relative to $n$.

$\nabla \times F=(3y^2,3x^2,3x^2+3y^2)$, whose divergence is $0$.

Take the outward normal on the cylindrical side and the upward normal on the top disc. The induced boundary direction on the bottom circle is clockwise as viewed from above.

The clockwise boundary integral at $z=0$ is
$-{\int }_0^{2\pi }{a}^4({\sin }^4\varphi +{\cos }^4\varphi )d\varphi =-3\pi a^4/2$. Direct integration of the curl over the side and top gives the same value (the $h$-dependent terms cancel), verifying Stokes.

$S\prime$ has the same bottom boundary and induced orientation, so Stokes’ theorem makes the answer independent of its shape: $-3\pi a^4/2$.