Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V resit / Question 3

Let $\Sigma$ be a compact, piecewise smooth oriented surface with piecewise smooth boundary, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of $\Sigma$. The Stokes theorem statesThe boundary orientation must agree with the chosen unit normal vector by the right-hand rule. For an upward orientation on a horizontal disk, its boundary is traversed counterclockwise when viewed from above. Reversing the orientation reverses both the line integral and the surface integral.

Choose the outward unit normal vector on the upper hemisphere. The curl of $\mathbf{F}=(4y,x,2z)$ isIn spherical polar coordinates, putHere $n_z=\cos \phi$ and $dS=a^2\sin \phi d\phi d\theta$. The surface integral is thereforeThe compatible boundary orientation follows increasing $t$ in $\mathbf{r}(t)=(a\cos t,a\sin t,0)$, $0\le t\le 2\pi$. On this circle,Thus the line integral iswhich agrees with the surface integral as required by the Stokes theorem.

Use the outward orientation on the unit hemisphere and counterclockwise boundary orientation on the unit circle. With $\mathbf{r}(t)=(\cos t,\sin t,0)$ and $f=x=\cos t$, the vector line integral isFor the surface integral, $\nabla f=\mathbf{i}$ and the outward unit normal vector is $\mathbf{n}=(x,y,z)$, since the radius is one. The cross product isThe integral of $y$ is zero by reflection symmetry. Using the same spherical polar coordinates as in ii, with $a=1$, the other component isHenceverifying the vector-valued Stokes identity directly, component by component. The wedge symbol in the paper denotes the cross product here.

Choose the scalar fieldThe allowed vector-valued Stokes identity givesEvery boundary point lies on the unit circle in the $xy$-plane, so $f=1/2$ throughout the boundary. Since the line integral of $d\mathbf{r}$ around a closed curve is the zero displacement,Only the boundary value of this scalar field mattered: the interior of the surface need not be part of a sphere. Either consistent choice of orientation gives the same zero answer.