Does zero curl imply a conservative field? A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9).
 A: You have to keep in mind that a vector field is not just a set of functions, but also a domain.  For instance, the vector field $\mathbf{F} = \left<-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2}\right>$ on the set $U = \left\{(x,y) \neq (0,0)\right\}$ has a curl of zero.  But it's not conservative, because integrating it around the unit circle results in $2\pi$, not zero as predicted by path-independence.
On the other hand, the same vector field restricted to $U' = \left\{x>0\right\}$ is conservative.  A potential function is $f(x,y) = \arctan\left(\frac{y}{x}\right)$.
The difference is that $U'$ is simply connected, while $U$ is not.  In fact, this is an symbolic version of M. Winter's graphical example.
A: Any conservative vector field $F :U \to \mathbb{R}^3$ is irrotational, i.e. $\mathbf{curl} (F)=0$, but the converse is true only if the domain $U$ is simply connected (see here for a classical example).
A: Not necessarily. Look at the following potential $A%$ defined on some region:

The associated vector field $F=\mathrm{grad}(A)$ looks like this:

Since it is a gradient, it has $\mathrm{curl}(F)=0$. But we can complete it into the following still curl-free vector field:

This vector field is curl-free, but not conservative because going around the center once (with an integral) does not yield zero.
This happens because the region on which $F$ is defined is not simply connected (i.e. it has a hole). If you are only interested in vector fields on all of $\Bbb R^3$, then you are safe: $\Bbb R^3$ is simply connected and every curl-free vector field is conservative.
