My question is really ''what is the definition of a conservative vector field''?
I've consulted 3 textbooks that all say a vector field $\vec{F}$ is conservative by definition if there exists a scalar potential $\phi$ such that $\nabla \phi = \vec{F}$. Then, they go on to talk about connected domains, path independence and the equality of mixed partials and how they are all related.
In particular, they emphasize that e.g. in $\mathbb{R}^2$ given $\vec{F} = \bigl<F_1,\,F_2\bigr>$, if $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$ on a simply-connected domain, then $\vec{F}$ is conservative on that domain.
However, without fail, all of them then offer the example of $\vec{F} = \bigl< \frac{-y}{x^2+y^2},\, \frac{x}{x^2+y^2} \bigr>$, pointing out:
- it's line integral is not path independent, even though
- $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$
and this is explained by pointing out that the domain is not simply connected (if the path contains the origin).
That much makes sense to me: we cannot conclude $\vec{F}$ is conservative based on the partial derivatives, because the domain is not simply connected - totally consistent with what has been presented.
What none of them address is why $\vec{F} = \bigl< \frac{-y}{x^2+y^2},\, \frac{x}{x^2+y^2} \bigr>$ is not conservative when there exists a potential $\phi = \arctan(y/x)$ such that $\nabla \phi = \vec{F}$. None of the texts mention any necessary conditions on the scalar potential. So is the existence of a scalar potential the definition of a vector field being conservative or not?