= Solution
Use $\gamma(t)=(\cos t,\sin t,0)$ for $0\le t\le2\pi$. On this path $\mathbf F=(-\sin t,\cos t,0)=\gamma'(t)$, so its <line integral> is
$$
\boxed{\oint_\gamma\mathbf F\cdot d\mathbf x=\int_0^{2\pi}1\,dt=2\pi.}
$$
The 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 is
$$
\partial_x(x/q)-\partial_y(-y/q)
=\frac{y^2-x^2}{q^2}-\frac{y^2-x^2}{q^2}=0.
$$
Thus 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. \b[Zero <curl> alone does not force global path independence.]
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