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Let $g$ be a continuous $2\pi$-periodic function and $\mathbf{F}=g(\theta ){\mathbf{e}}_{\theta }/r$ on $r>0$. On a circle about the axis, $d\mathbf{r}=r{\mathbf{e}}_{\theta }d\theta$, giving the displayed line integral. A single-valued potential of a conservative vector field on the full punctured space exists if and only if this integral is zero. Necessity is the fundamental theorem for line integrals. For sufficiency, the primitive $P(\theta )={\int }_0^{\theta }g(s)ds$ is $2\pi$-periodic when the mean is zero, and the gradient in cylindrical coordinates gives $\nabla P=\mathbf{F}$.

On a domain with one angular cut removed, the same primitive gives a potential of a conservative vector field even when the mean is nonzero, because the angle has a single-valued continuous branch there. For example $g(\theta )=\sqrt{1+\cos \theta }$ has integral $4\sqrt{2}$ around the full circle but primitive $2\sqrt{2}\sin (\theta /2)$ on $-\pi <\theta <\pi$.

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.