Verify $\nabla \times \textbf{f} = 0$ and Show Non-Existence of $\psi$ on $R_2$

[rsq_a]

I wasn't quite sure how to do the second part of this question:

Given \textbf{f}(x,y,z) = (y/(x^2+y^2), -x/(x^2+y^2), 0) where (x,y) \neq (0,0), verify that \nabla \times f = 0.

(A) Find a scalar field \phi such that \textbf{f} = \nabla \phi on R_1 = \{(x,y,z): y > 0\}.

(B) Show that there does NOT exist \psi such that \textbf{f} = \nabla\psi on R_2 = \{(x,y,z): (x,y) \neq (0,0)For (A), I found \phi = arctan(x) + arccot(x) - arctan(y/x).

I'm not sure how to do (B). In fact, I'm not even sure why it's true.

 

[rsq_a]

Never mind. It's pretty easy to show just by considering a contour going around the pole (x,y) = (0, 0). Considering the path integral around r(t) = (cos t, sin t, 0), the integral gives 2\pi, whereas it should give 0 if a potential function did exist.