 Solution

Use $\gamma (t)=(\cos t,\sin t,0)$ for $0\le t\le 2\pi$. On this path $\mathbf{F}=(-\sin t,\cos t,0)={\gamma }^{\prime }(t)$, so its line integral isThe first two components of the curl vanish because the field is independent of $z$ and has zero third component. Writing $q=x^2+y^2$, its third curl component isThus this azimuthal inverse-radius vector field is a curl-free vector field on $G={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, but its line integral is not path independent. This domain is not simply connected: the circle links the removed axis. A spanning disk through the axis is inadmissible for Stokes theorem, since the vector field is undefined there. Zero curl alone does not force global path independence.