Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 9A (Vector Calculus)

Using ${\epsilon }_{ijk}{\epsilon }_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$,
$[\nabla \times (F\times G){]}_i={\partial }_j(F_i{G}_j)-{\partial }_j(F_j{G}_i)$; expanding the derivatives gives the stated four terms.

Irrotational means $\nabla \times E=0$. For $E=\nabla f$, its $i$th curl component is ${\epsilon }_{ijk}{\partial }_j{\partial }_{k}f=0$ because mixed partials are symmetric.

On $C$, $dx=a{e}_{\varphi }d\varphi$, so $I={\int }_0^{2\pi }d\varphi =2\pi$. Locally $f=\varphi$, but no single-valued global potential exists on the punctured domain; this is why irrotationality does not force this closed integral to vanish.

The field is $\nabla f$ for $f=\rho \sin \varphi e^{-{\rho }^2-z^2}$. Hence $I=0$, either directly or by the fundamental theorem for line integrals.