Regarding a conservative field $\vec{F}$ in a region $D \subseteq R^2$, I know that the requirements are:
- Curl of $\vec{F}$ is $0$.
- $\vec{F}$ is defined in D (doesn't have singularities in D).
But I have doubts about the conditions that must be met for a vector field to be considered conservative in $R^3$, as I have understood (by the comments of teachers or test solutions I've seen), it is not necessary to show that a vector field is defined in a region $D$ to consider it as conservative in D, it seems that that is enough if the curl is 0, but I never managed to understand it (if it really is like that), why is that so?
The correct condition is that the field to be defined on a simply connected subset of $\mathbb R^n$ before a local condition is enough to be sure it is conservative.
For example, consider the following field defined on $\mathbb R^3$ except for the $z$-axis: $$ \vec F(x,y,z) = \frac1{x^2+y^2}(-y,x,0) $$ This is not conservative -- if you go around the unit circle in the $xy$-plane you either win or lose $2\pi$ energy, depending on the direction. But the curl is $0$ everywhere the field is defined.