A curl-free $C^1$ field on the complement of a line that is not conservative

The field is $C^1$ on $U$, because the denominator $x^2+y^2$ never vanishes there and the coordinate functions are rational in $x$ and $y$.

The first two curl coordinates vanish because $F_z=0$ and the first two components do not depend on $z$, while the third is ${\mathrm{\partial }}_x(x\mathrm{/}(x^2+y^2))-{\mathrm{\partial }}_y(-y\mathrm{/}(x^2+y^2))=0$ by the quotient rule and cancellation. Therefore $\mathrm{curl}F=0$ on $U$.

On the unit circle $\gamma (t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$, one has $F(\gamma (t))=(-\sin t,\cos t,0)={\gamma }^{\mathrm{\prime }}(t)$ by [L5] and [L6], so ${\int }_{\gamma }F\cdot d\mathbf{r}={\int }_0^{2\pi }1dt=2\pi$ by [F3], [F6], and [L7].

By [L1], the field is closed on $U$.

If $F$ were conservative, [L3] and [F2] would force the closed-loop integral in step 2.2 to be $0$, a contradiction. Hence $F$ is not conservative.

The domain is open, connected, and not star-shaped. To see openness, fix $p=(x,y,z)\in U$ and put $d=\frac{1}{2}\max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}>0$. Every point of the deleted axis differs from $p$ by at least $2d$ in one of its first two coordinates, so the Euclidean ball $B(p,d)$ misses that axis and lies in $U$. To see connectedness, use [L9] to write $(x,y)=r(\cos \theta ,\sin \theta )$ with $r>0$. The circular arc $s\mapsto (r\cos ((1-s)\theta ),r\sin ((1-s)\theta ),z)$ joins $(x,y,z)$ to $(r,0,z)$, the radial segment $s\mapsto ((1-s)r+s,0,z)$ joins that point to $(1,0,z)$, and the vertical segment $s\mapsto (1,0,(1-s)z)$ joins it to $(1,0,0)$; all three pieces are piecewise $C^1$ and stay in $U$. Thus $U$ is path-connected and hence connected by [L8]. Finally, for any proposed star centre, the segment to its reflection across the deleted axis meets that axis, so $U$ is not star-shaped. This is exactly the hypothesis of [L2] that fails.