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

In Cartesian components, $\mathbf{F}=(z-y,x,-x)$, soIt cannot be a conservative vector field. For the second field, put $\Phi (\mathbf{r})=-(1+|\mathbf{r}{|}^2{)}^{-1/2}$. Its gradient is $\nabla \Phi =\mathbf{r}/(1+|\mathbf{r}{|}^2{)}^{3/2}=\mathbf{G}$. This smooth global scalar function proves $\mathbf{G}$ is conservative, with $\boxed{\nabla \times \mathbf{G}=0}$. The following path integrals exhibit the different behavior directly.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$. Its tangent is $(-1,1,0)$ and $\mathbf{F}=(-t,1-t,t-1)$, so $\mathbf{F}\cdot {\mathbf{r}}^{\prime }=1$. Hence its line integral isFor the other line integral, $|\mathbf{r}{|}^2=(1-t{)}^2+t^2$ and $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2t-1$. ThusThis also equals the change in $\Phi$, by the fundamental theorem for line integrals.

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.