Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 13

Split the vector field into $a(-y,x,0)$ and $e^{-r^2}(x,y,z)$. The second part is the gradient of $-\frac{1}{2}{e}^{-r^2}$, so its curl vanishes. The first part has only a vertical curl component, ${\partial }_x(ax)-{\partial }_y(-ay)=2a$. Hence

The curl found in part (a) is constant, so its divergence isThis also follows from the identity divergence of a curl is zero, applicable here because the field is smooth everywhere.

By part (a), the vector field is irrotational exactly when $2a=0$. Thus

The original PDF has angle $\theta (t)=2\pi t^n$, with $n$ an exponent on $t$. It is not $2\pi nt$: each permitted value of $n$ changes the speed but traverses the same circle once counterclockwise. Along the path, $r^2=2$ andThe radial contribution to the line integral vanishes, since $(\cos \theta ,\sin \theta ,1)\cdot (-\sin \theta ,\cos \theta ,0)=0$. The rotational contribution givesThereforeThe answer is independent of the monotone parametrization, as an oriented line integral should be.

The upward unit vector extracts the vertical curl component:It is constant over the unit disc, soThis agrees with part (d) by Stokes theorem: the boundary circle has counterclockwise orientation when viewed from the upward normal, including for the parametrizations $2\pi t^n$.

For $a=0$, differentiate the proposed scalar potential directly:ThusFor $a\ne 0$, the curl is the nonzero vector $(0,0,2a)$. A gradient of a twice differentiable scalar field has zero curl, so no such scalar field exists. Equivalently the closed line integral in part (d) is nonzero, contradicting the path independence of a conservative vector field. The field is smooth on all of ${\mathbb{R}}^3$, so no excluded singular set complicates this conclusion.