 Solution

Away from the axis, cylindrical coordinates put the field in the line vortex formThe only potentially nonzero component of its curl isThe other components vanish because there is no axial dependence and no radial or axial velocity. Thus the curl is zero for $\rho >0$; the field is singular on the axis, so a classical curl there is not defined.

For a spanning surface avoiding that axis, the smoothness hypotheses of Stokes theorem hold and give ${\oint }_C\mathbf{v}\cdot d\mathbf{r}=0$. If the enclosed planar region contains the axis, the singularity prevents applying that theorem directly. Excise a small disk about the origin. The resulting annulus has zero curl, so its outer circulation equals the counterclockwise circulation on the small circle:This sign assumes positive, counterclockwise orientation. Clockwise orientation gives $-\kappa$; more generally the value is $\kappa$ times the winding number about the origin. A curve through the origin is excluded because its integral is singular. In the distribution interpretation the missing vorticity is concentrated on the axis as $\kappa \delta (x)\delta (y){\mathbf{e}}_z$, explaining the nonzero circulation despite zero classical curl away from it.