Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 16

The first field is ${\mathbf{F}}_1=(z-y,x,-x)$, so direct curl calculation givesIt is not conservative because its curl is nonzero. The second is a radial gradient field:The scalar expression is a smooth function and single-valued on all of space, so ${\mathbf{F}}_2$ is a conservative vector field. Its potential of a conservative vector field is $\Phi =-(1+r^2{)}^{-1/2}$. Parts (a) and (b) check path dependence directly.

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=(-1,1,0)$. For the first field, ${\mathbf{F}}_1=(-t,1-t,-1+t)$ and its dot product with the tangent is one. ThereforeFor the second field, use the fundamental theorem for line integrals. The endpoints both have $r=1$, so their potential values agree andDirectly, its parameter integrand is $(2t-1)/[1+(1-t{)}^2+t^2{]}^{3/2}$, which changes sign under $t\mapsto 1-t$ and integrates to zero.

The shorter circular arc lies in the $x$-$y$ plane and can be written $\mathbf{r}(\theta )=(\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. The first field is $(-\sin \theta ,\cos \theta ,-\cos \theta )$, whose dot product with ${\mathbf{r}}^{\prime }$ is one. HenceThe second field is radial while the circular tangent is perpendicular to the radius, so $\boxed{{\int }_{\mathrm{arc}}{\mathbf{F}}_2\cdot d\mathbf{r}=0}$. Thus the first field gives different values, $1$ and $\pi /2$, along the two paths, whereas the conservative second field gives the same endpoint difference.

As a check using Stokes theorem, the arc followed by the reversed segment encloses area $\pi /4-1/2$ in the plane. The constant curl has upward component two, giving circulation $2(\pi /4-1/2)=\pi /2-1$, exactly the difference above.