If the integral of a vector field over a closed curve equals zero, is the field conservative? If a vector field has a potential, then the integral of that vector field over every closed curve is zero. If the integral of a vector field over a closed curve equals zero, does that imply that the vector field is conservative?
 A: It's certainly not sufficient for the condition to hold for some path; for one, the line integral of any vector field along a constant path is $0$, but not every vector field is conservative.
On the other hand, we have:

Theorem A vector field $\bf F$ (on say, some open set) is conservative iff the line integral of a vector field $\bf F$ over every closed curve in the domain of $\bf F$ is $0$.

The forward implication is a consequence of the F.T.C. for line integrals. We can prove the converse by constructing an explicit potential $f$: If we fix a point $p$ in the domain, for every point $q$ define
$$f(q) := \int_{\gamma} {\bf F} \cdot d{\bf s},$$
where $\gamma$ is any path from $p$ to $q$; this is well-defined (i.e., independent of the path $\gamma$ chosen) because the integral over a closed curve is zero. All that remains to check is that $\nabla f = {\bf F}$ as claimed (see Independepnce of path in a closed curve line integral).
A: In differential form language, one can view one side of this from a perspective given by Stokes theorem. The vector field proxy being conservative is equivalent to the corresponding differential form $\omega$ being exact, so it is closed $d\omega=0$. Then, since $\gamma$ is a closed curve, it bounds a domain $\Omega$ such that $\gamma=\partial \Omega$, and
$$ \int_{\gamma}\omega = \int_{\Omega} d\omega=0. $$
We see that the necessary assumption is closedness, so in fact that $\omega$ is exact is more than what is required since not every closed form is exact.
A nice exercise is to see that the integral of any of the three canonical basis vector fields $x_1e_1$,$x_2e_2$ and $x_3e_3$ over a closed curve is 0. One can take a triangle for instance, to easily calculate the left hand side of Stokes theorem. Indeed, for each of these $d\omega=0.$
