Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 1 / 9 / ii / Solution

On each of the two radial legs, $\mathbf{F}=\mathbf{a}\times \mathbf{r}$ is perpendicular to the tangent. Each contribution is zero, giving $\boxed{\int \mathbf{F}\cdot d\mathbf{r}=0}$. This differs from the straight-path answer, confirming path dependence.

For $\mathbf{G}$, use $\mathbf{r}=(s,0,0)$ with $s$ decreasing from one to zero, followed by $\mathbf{r}=(0,s,0)$ with $s$ increasing from zero to one. The two contributions areThey cancel: $\boxed{\int \mathbf{G}\cdot d\mathbf{r}=0}$. This agreement is a consequence of the conservative vector field property already proved, rather than evidence sufficient by itself to establish it.