Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 3

Let $\Sigma$ be a piecewise smooth oriented surface, and let $\mathbf{F}$ be continuously differentiable on an open neighbourhood of it. Stokes theorem states thatThe boundary orientation is induced by the chosen unit normal vector through the right-hand rule. Smoothness on a neighbourhood makes the spatial partial derivatives in the curl well defined.

Choose the outward orientation of a surface on the lower half of the ellipsoid. At its lowest point the normal points down, so its boundary ellipse is traversed clockwise when viewed from above. UseThe line integral is

For an independent surface integral calculation, first compute $\nabla \times \mathbf{F}=(2y,-2x,1)$. Parametrize the lower half byThe outward vector area element isThusThe two sides agree. Choosing the inward unit normal vector and its corresponding counterclockwise boundary orientation changes both answers to $6\pi$.

The sufficiency claim requires a domain qualification. For a continuously differentiable vector field on all of ${\mathbb{R}}^3$, or more generally on a simply connected open domain, the equivalence is valid. On an arbitrary domain, vanishing curl is necessary but may fail to be sufficient.

Necessity. Suppose every closed-curve line integral vanishes. Fix a point $p$ in the open domain and a unit vector $a$. A sufficiently small planar disk $D_{\epsilon }$ of radius $\epsilon$, centred at $p$ and normal to $a$, lies in the domain. Stokes theorem givesDivide by $\pi {\epsilon }^2$ and let $\epsilon \downarrow 0$. By continuity, the average tends to $(\nabla \times \mathbf{F})(p)\cdot a$. Taking the three coordinate directions proves $\boxed{\nabla \times \mathbf{F}=0}$ everywhere.

Sufficiency on ${\mathbb{R}}^3$. A direct proof constructs a potential of a conservative vector field. PutWrite $F_j$ for the components of $\mathbf{F}$. Vanishing curl says ${\partial }_i{F}_j={\partial }_j{F}_i$. Differentiation under the integral, followed by the chain rule, givesThis is the radial integral potential for a curl-free field. The fundamental theorem for line integrals now gives zero on every closed curve. On a simply connected open domain the analogous argument uses path integrals: every loop can be contracted within the domain, and applying Stokes theorem to a piecewise smooth contraction makes its circulation zero. Equivalently, whenever a particular loop bounds an oriented spanning surface entirely in the domain, Stokes theorem immediately gives zero for that loop.

Why the domain matters. On $U={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, the smooth vector fieldhas zero curl, but its line integral around the unit circle in $z=0$ is $2\pi$, as in ii. No spanning surface for this loop can lie entirely in $U$. The literal claim without a condition on the domain therefore has this counterexample.

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.