 Solution

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.