Doesn't the existence of non-conservative fields with zero curl contradict the Stokes Theorem?

Question

I just saw an exemple of a non-conservative field with zero curl. It is $$f(x,y)=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right).$$ However the classical Kelvin-Stokes theorem says that $$\oint_{\gamma} F\cdot\mathrm{d}r=\iint_S(\nabla\times F)\cdot\mathrm{d}S.$$ Because $\nabla\times F=0$, we have that the integral on the right is zero. Doesn't that imply that the integral on the left if zero too?

Answer 1

The Kelvin-Stokes theorem is valid for a smooth vector field, but your field is not defined for $(x,y)=(0,0)$ so it is not smooth in a region that contains the origin.