Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 1

Here $\varphi$ is a scalar field: the source's description as a vector field is a typographical slip, since the displayed formula is for its gradient. For cylindrical coordinates, write $x=r\cos \theta$, $y=r\sin \theta$. Away from $r=0$, a local choice of $\theta$ gives the partial derivativesThe chain rule therefore givesUse the Cartesian coordinates definition $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ and the unit vectorsCollecting their coefficients provesThis is the gradient in cylindrical coordinates. The factor $1/r$ converts a change in angle into a change per unit arc length.

In cylindrical coordinates, $(-y,x,0)=r{\mathbf{e}}_{\theta }$ and $x+r=r(1+\cos \theta )$. Hence the vector field simplifies toTake the orientation of $C$ to be increasing $\theta$, as in its given parametrization. On the circle, $d\mathbf{r}=a{\mathbf{e}}_{\theta }d\theta$, so its line integral isWe used $\sqrt{1+\cos \theta }=\sqrt{2}|\cos (\theta /2)|$; the cosine is nonnegative on this interval. In particular, the answer does not depend on $a$.

Here a potential of a conservative vector field means a single-valued scalar field $\psi$ with $\nabla \psi =\mathbf{F}$. By the fundamental theorem for line integrals, its line integral around every closed curve must vanish. Here the angular circulation obstruction to a cylindrical potential lets us test this using a circle about the axis.

On $V_1$, the circle from i lies entirely in the region and has line integral $4\sqrt{2}$. Therefore there is no potential of a conservative vector field on $V_1$.

On $V_2$, choose the continuous angular branch $-\pi <\theta <\pi$. By a, the equation for a potential of a conservative vector field reduces to ${\psi }_r={\psi }_z=0$ and ${\psi }_{\theta }=\sqrt{2}\cos (\theta /2)$. ThusThe denominator is positive throughout $V_2$, including when $y=0$ and $x>0$, so this is a well-defined differentiable potential of a conservative vector field on the entire region.

For $V_3$, choose the particular circle $x=2\cos \theta$, $y=2\sin \theta$, $z=0$. Its only point on the removed half-plane's supporting cut is $(-2,0,0)$, where $(x+2{)}^2+z^2=0<1$; this point is retained by $V_3$. Thus the whole circle lies in $V_3$ and again has line integral $4\sqrt{2}$. Consequently there is no potential of a conservative vector field on $V_3$. The opening in the cut allows a closed curve to go around the excluded axis.

Finally $V_4\subset V_2$. Restricting the potential of a conservative vector field already constructed on $V_2$ givesThe extra excluded cylinder does not invalidate an existing potential of a conservative vector field. Thus being simply connected is a sufficient condition in the usual curl-free criterion, not a necessary condition for a particular vector field to have a potential of a conservative vector field.