Past exam of the mathematical methods II course of the University of Oxford / 2016 / 8

For a continuously differentiable vector field $\mathbf{F}$ near a piecewise smooth oriented surface $S$, Stokes theorem statesThe circulation around the whole boundary equals the flux of the curl through the surface. The unit normal vector and boundary direction must be compatible with the right-hand rule: curling the right-hand fingers in the boundary direction points the thumb along the normal. Reversing both orientations reverses both integrals.

Apply Stokes theorem to $\mathbf{F}=\psi \mathbf{B}$, with $\mathbf{B}$ an arbitrary constant vector. The curl product rule for a scalar field reduces to $\nabla \times (\psi \mathbf{B})=\nabla \psi \times \mathbf{B}$. Using the scalar triple product,Since these scalar products agree for every $\mathbf{B}$, the vectors agree. This proves the vector-valued Stokes identityIn particular, the ordering of the cross product is normal crossed with gradient; reversing it would change the sign.

By Stokes theorem, it suffices to calculate the boundary line integral. On the $xz$ plane, $y=0$ and $dy=0$, soThis exact differential has zero integral around a closed curve. ThereforeThe argument applies to any admissible spanning surface, including a curved one; the surface itself need not lie in the $xz$ plane.

Choose the oriented surface parametrization $\mathbf{r}(x,y)=(x,y,1-x)$, with $0\le x,y\le 1$. Then ${\mathbf{r}}_x\times {\mathbf{r}}_y=(1,0,1)$ and $\nabla \times \mathbf{A}=(-3z,x^2,y)$. The surface integral isFor the compatible boundary direction, use successive vertices $(0,0,1)$, $(1,0,0)$, $(1,1,0)$, $(0,1,1)$ and back. The four line integrals are respectivelyTheir sum is $-1$, agreeing with the surface integral. Choosing the opposite normal and reversing the boundary direction gives $+1$ on both sides. Stating an orientation is necessary because the question does not prescribe one.