Explain why $F(x,y) = \Big(\frac{-y}{x^2 + y^2}, \frac{x}{x^2+y^2}\Big)$ cannot be the gradient of a function (defined away from the origin). Can it be the gradient if we only require F and $f$ to be defined away from the y-axis?
My thoughts: So my initial thought is that we can back out $f$ to find $f(x,y) = -arctan(\frac{x}{y})$ and that is undefined at $y=0$.. but I don't think that in itself disqualifies it as a gradient.
Function looks like this, in case you're curious: link
The domain of the field $F$ is the plane with the origin removed, note this is not simply connected. Since the domain of $F$ is not simply connected you cannot conclude that $F$ must be conservative solely on the basis of the cross-partial criterion (i.e. curl$(F) = 0$).
To see why $F$ is not conservative consider the line integral of $F$ across the unit circle parameterized by $r(t) = < \cos(t), \sin(t) >$. Then $F(c(t)) \cdot c'(t) = 1 $, thus the line integral evaluates to $2\pi$. If $F$ were conservative then the line integral (or circulation) across any closed curve would be $0$.
However, even though the field isn't conservative on its domain, it is conservative in any smaller, simply connected domain like the upper half-plane, $y > 0$. For instance, in this upper half plane you can take $f(x,y) = \arctan(y/x)$ as a potential function for $F$.