Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 4

For a sufficiently smooth vector field on an oriented piecewise-smooth surface $S$ with boundary $\partial S$, Stokes theorem statesHere $\mathbf{n}$ is the chosen unit normal, $dS$ is the scalar surface-area element, and $d\mathbf{r}$ follows the positively oriented boundary. The boundary orientation is fixed by the right-hand rule: the fingers curl along $\partial S$ when the thumb points along $\mathbf{n}$.

The Cartesian curl is obtained by cross-differentiating the components:

No scalar potential exists on the full three-dimensional domain. A sufficiently smooth gradient field has zero curl, since mixed partial derivatives commute. The calculated curl is not identically zero. The fact that it vanishes at some individual points does not make the whole field a gradient field.

Orient the triangle by the path $A\to B\to C\to A$. The cross product $(B-A)\times (C-A)=(0,1,0)$ gives the positive normal vector. In the plane $y=1$, the triangular region is $x,z\ge 0$, $x+z\le 1$. The surface integral in Stokes theorem is thereforeFor the boundary line integral, $AB$ has $z=0$, so contributes zero. On $BC$, $x=0$ and only $dz$ is nonzero, so its contribution is also zero. Parametrize $CA$ by $(x,y,z)=(t,1,1-t)$, $0\le t\le 1$. Its contribution isThus both oriented integrals equal $1/15$. Reversing the orientation would reverse both signs.