Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 4A (Vector Calculus)

Parametrize the smooth path by $\gamma :[0,1]\to G$. The chain rule turns the line integral of the gradient into a one-variable derivative:This is the fundamental theorem for line integrals; in particular the integral has path independence.

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:A smooth gradient field is necessarily curl-free.

A sufficient condition is that $G$ be an open simply connected domain, with $\mathbf{F}$ continuously differentiable. Under this condition a curl-free vector field has a global potential of a conservative vector field and hence path independence by the fundamental theorem for line integrals. One way to see the global step is to deform the closed loop formed by two paths into a point inside $G$: the integral is unchanged during the deformation because the curl is zero, as expressed by Stokes theorem. Therefore simple connectivity is a sufficient domain hypothesis. It is not a claim that every particular curl-free field needs such a domain; a given field can have a global potential on a domain with holes.

Use $\gamma (t)=(\cos t,\sin t,0)$ for $0\le t\le 2\pi$. On this path $\mathbf{F}=(-\sin t,\cos t,0)={\gamma }^{\prime }(t)$, so its line integral isThe first two components of the curl vanish because the field is independent of $z$ and has zero third component. Writing $q=x^2+y^2$, its third curl component isThus this azimuthal inverse-radius vector field is a curl-free vector field on $G={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, but its line integral is not path independent. This domain is not simply connected: the circle links the removed axis. A spanning disk through the axis is inadmissible for Stokes theorem, since the vector field is undefined there. Zero curl alone does not force global path independence.