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.