Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 2

For a continuous vector field along an oriented piecewise continuously differentiable parametric curve $\gamma :[a,b]\to {\mathbb{R}}^3$, define its line integral byadding the integrals over the smooth pieces when necessary. The value refers to the specified traversal: going around a curve twice counts twice.

An orientation-preserving reparametrization has the form $\tilde{\gamma }(s)=\gamma (h(s))$, where $h:[\alpha ,\beta ]\to [a,b]$ is an increasing, piecewise smooth change of parameter with $h(\alpha )=a$, $h(\beta )=b$. For regular parametrizations one may take $h$ and its inverse continuously differentiable on each piece. The chain rule gives ${\tilde{\gamma }}^{\prime }(s)={\gamma }^{\prime }(h(s))h^{\prime }(s)$. The change of variables $t=h(s)$ therefore yieldsThis proves reparametrization invariance of a vector line integral. If $h$ is decreasing, the endpoints in the final integral are reversed and the sign changes. Thus the same direction and the same multiplicity of traversal are essential.

For the parametric curve $\mathbf{r}(t)=(t,t^2,t^3)$, $0\le t\le 1$, the vector field and the tangent vector areConsequently the line integral is

Compute the line integral on each straight segment. On the first, $y=z=0$ and only $dx$ contributes; on the second, $x=1,z=0$ and only $dy$ contributes; on the third, $x=y=1$ and only $dz$ contributes. ThereforeAlthough $F_x=9$ on the second segment, that term does no work there because $dx=0$.

Parametrize the straight segment by $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$. Then the line integral isThe three different answers are possible because this vector field is not a conservative vector field; for example ${\partial }_y{F}_x=6$ whereas ${\partial }_x{F}_y=0$.

Take counterclockwise traversal as viewed from $z>0$. In the plane $z=0$, the work is the planar line integral of $P=2x-y$ and $Q=x+y$. Green's theorem givesThe $z$ component contributes nothing because $dz=0$. The question specifies one circuit without specifying its direction: clockwise traversal gives $\boxed{-18\pi }$.