Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 11B (Vector Calculus)

In components,because the Levi-Civita symbol is antisymmetric in $j,k$ while equality of mixed partial derivatives makes ${\partial }_j{\partial }_k\varphi$ symmetric. Hence

Similarly,since the last derivatives are symmetric in $i,j$. Thus

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

The first field isso its integral around $C$ is zero and it is conservative with potential $e^{x^2}\cos (yz)$.

Along $C$, $dr=(-\sin s,\cos s,0)ds$ andsoThus $B_2$ is not conservative and cannot be a gradient on the stated punctured region.

Finally,Its integral around $C$ is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectivelyand exactly $B_1,B_3$ are conservative gradient fields.