= Solution
{c}
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> $4\sqrt2$. Therefore \b[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 $\psi_\theta=\sqrt2\cos(\theta/2)$. Thus
$$
\boxed{\psi=2\sqrt2\sin(\theta/2)+C
=\frac{2y}{\sqrt{r(r+x)}}+C,\qquad r=\sqrt{x^2+y^2},\quad\text{on }V_2.}
$$
The 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> $4\sqrt2$. Consequently \b[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 $V_2$ gives
$$
\boxed{\psi=\frac{2y}{\sqrt{r(r+x)}}+C\quad\text{on }V_4.}
$$
The 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>.
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