Show that a vector field is not conservative (example) Let $\Omega=\mathbb{R^2}\smallsetminus\{(0,0)\}$ and $$\vec{F}(x,y)=-\frac{y}{x^2+y^2}\vec{i}+\frac{x}{x^2+y^2}\vec{j}$$ To show that the vector field $F$ is not conservative $$\vec{\nabla}\times \vec{F}=0\,\vec{i}+0\,\vec{j}+\bigg(\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}\bigg)\vec{k}=\vec{0}$$ which indicates that it is conservative(?). Alternatively, the scalar function
$$f(x,y)=-\arctan\bigg(\frac{x}{y}\bigg)$$ is the potential function of the field $F$. If $$\int_C\vec{F}\cdot\mathrm{d}\vec{r}\neq f\big(x(\beta),y(\beta)\big)-f\big(x(\alpha),y(\alpha)\big)$$ $\big($where $r(t)$ is the parametrization of an arbitrary curve $c$ and $\alpha$ and $\beta$ are its starting and end point respectively$\big)$ 
then the fundumental theorem for line integrals is not satisfied and thus $F$ is not conservative.
I tried using $r(t)=t\vec{i}+t\vec{j}, \space t\in[\alpha,\beta]$, but it didn't work.  
What curve would be a better choice for $C$ and what's the deal with $\mathrm{rot}\,F$ being zero?
 A: One of the most important "functions" in classical analysis is the ${\rm arg}$ function, giving the polar angle of a point $(x,y)$, resp., of $z=x+iy$,  in the punctured plane "up to an additive constant $2k\pi$". Locally, i.e., in suitable neighborhoods of points $(x_0,y_0)\in\dot{\mathbb R}^2$ this function has well-defined real representants. E.g., in the half plane $x>0$ we may choose the representant $$\phi(x,y):=\arctan{y\over x}\ .$$
Now this ${\rm arg}$ "function" has a well-defined gradient
$$\nabla{\rm arg}(x,y)=\left({-y\over x^2+y^2},{x\over x^2+y^2}\right)\qquad\bigl((x,y)\ne(0,0)\bigr)\ .$$It turns out that your ${\bf F}$ is nothing else but $\nabla{\rm arg}$. Since, locally,  ${\rm arg}$ is represented by smooth functions $(x,y)\mapsto\phi(x,y)$ it follows that ${\rm curl}\,{\bf F}\equiv0$.
But it is impossible to concatenate the local polar angles $\phi(x,y)$ to one single real-valued polar angle function $\phi:\>\dot{\mathbb R}\to{\mathbb R}$. This then implies that the given ${\bf F}$ is not conservative. A definitive proof of this fact comes from computing $\int_\gamma {\bf F}\cdot d{\bf z}$ along the unit circle $\gamma$. Along this circle the argument increases continuously by $2\pi$, hence the value of this integral is $2\pi\ne0$.
A: The assertion that $\nabla \times {\bf F} = {\bf 0}$ is a necessary condition, not a sufficient one. (Note that this condition implicitly uses the unstated realization of $\Bbb R^2$ as the $xy$-plane in $\Bbb R^3$.) As Batominovski points out in the comments, this is the key lesson of the problem.
Hint Now, the scalar function $f(x, y) = -\arctan\left(\frac{x}{y}\right)$ is not a potential for $\bf F$---for the simple reason that $f$ is not defined on all of the domain of $\bf F$ (it isn't defined anywhere on the $x$-axis). On the other hand, it is a potential for the restriction ${\bf F}|_R$ of $\bf F$ to the right half-plane $R$, so that restriction is conservative, and hence any counterexample cannot use curves wholly contained within $R$.
It is, by the way, sufficient to think about closed curves $C$, since any two curves with the same endpoints form a closed loop. Then for any such curve and conservative vector field $\bf G$, your formula gives $\int_C {\bf G} \cdot d{\bf s} = 0$.
So: What is a closed curve---again not contained in $R$---along which it is easy to evaluate $\int_C {\bf F} \cdot d{\bf s}$ directly? If we can find a $C$ for which the integral is nonzero, the contrapositive of the previous observation gives that $\bf F$ is not conservative.

Additional hint One such curve is the unit circle, $x^2 + y^2 = 1$, on which ${\bf F} = -y {\bf i} + x {\bf j}$.

A: If a differentiable field $\vec{F}(x,y)=F_1(x,y)\vec{i}+F_2(x,y)\vec{j}$ is conservative then you must have $\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$ in every point of the domain. You can easily check that your field doesn't have this property. 
