Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V / 1

For an oriented surface $\Sigma$ with piecewise smooth closed boundary $\partial \Sigma$, and a vector field $\mathbf{F}$ that is continuously differentiable on an open neighborhood of $\Sigma$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal vector. The orientation of $\partial \Sigma$ follows the right-hand rule: with the right thumb pointing along $\mathbf{n}$, the fingers curl in the positive direction. Equivalently, walking along the boundary with one's head along $\mathbf{n}$ leaves the surface on the left. Each boundary component receives this induced orientation.

Write $r=\sqrt{x^2+y^2}>0$. The cylindrical coordinates give ${\mathbf{e}}_{\theta }=(-y/r,x/r,0)$ and ${\mathbf{e}}_r=(x/r,y/r,0)$, so the vector components of the vector field areThe first two components of the curl, calculated in Cartesian coordinates, are $-x{f}^{\prime }(z)/r$ and $-y{f}^{\prime }(z)/r$. For the third component, the chain rule givesand henceCombining these vector components in the cylindrical basis yieldsThe restriction $r>0$ matters because the cylindrical basis is undefined on the axis.

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.