Why does curl($F$)=$0$ $\iff$ $F$ is conservative? Why is it true that$$\displaystyle curl (\vec{F})=0 \iff \vec{F} \text{ is conservative}$$ i.e. $$\displaystyle \exists f \text{ s.t. }\nabla f=\vec{F}$$
 A: Assume for simplicity that the domain  $\Omega\subset{\mathbb R}^3$  of the vector field ${\bf F}$ is convex and contains the origin ${\bf 0}$, and consider a tiny parallelogram $P$ centered at some point ${\bf p}\in\Omega$. Its boundary $\partial P$ consists of four vectorial segments ${\bf U}$, ${\bf V}$, $-{\bf U}$, and $-{\bf V}$. Doing the calculation one finds that
$$\int_{\partial P}{\bf F}\cdot d{\bf r}={\rm curl}\,{\bf F}({\bf p})\cdot({\bf U}\times{\bf V})\ +o\bigl({\rm area}(P)\bigr)\ .$$
In particular, if ${\rm curl}\,{\bf F}\equiv{\bf 0}$, for any given $\epsilon>0$ one has 
$$\left|\int_{\partial P}{\bf F}\cdot d{\bf r}\right|\leq\epsilon\>{\rm area}(P)\tag{1}$$
for all sufficiently small parallelograms $P\subset\Omega$. It follows that in fact
$$ \int_{\partial R}{\bf F}\cdot d{\bf r}=0\tag{2}$$
even for  "large" rectangles $R$ aligned to the  coordinate axes, because such an $R$ can be subdivided into a million small $P$'s for which $(1)$ holds, and the contributions along all inner boundary edges cancel.
We now define a potential $f:\>\Omega\to{\mathbb R}$ as follows: For any point ${\bf x}\in\Omega$ let $\gamma({\bf x})$ be a path connecting ${\bf 0}$ with ${\bf x}$, and consisting of three line segments parallel to the coordinate axes. Then put
$$f({\bf x}):=\int_{\gamma({\bf x})}{\bf F}\cdot d{\bf r}\ .$$
From $(2)$ it follows that $f({\bf x})$ does not depend on the chosen $\gamma({\bf x})$ , and using a proper choice for each partial derivative of $f$ it is then easy to show that
$$\nabla f({\bf x})={\bf F}({\bf x})\qquad({\bf x}\in\Omega)\ .$$
