Here is a statement about conservative vector fields to be proved or disproved:
If $\vec{F}=(F1,F2) :\Omega \rightarrow \mathbb{R}^2\space$is a smooth vector field (here $\Omega\subseteq\mathbb{R}^2$) and $\frac{\partial{F1}}{\partial{y}}=\frac{\partial{F2}}{\partial{x}}\space$then $\vec{F}$ is conservative.
I think the following vector field defined on $\Omega=\mathbb{R}^2\backslash\{(0,0)\} $ might be a counter example against this statement, \begin{align} \vec{F}(x,y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right) \end{align} since its line integral along the unit circle centred at $(0,0)$ can be computed to be $2\pi$. But does it satisfy the smoothness supposition?
And why can't we apply the Green's theorem to yield that the line integral mentioned above is zero? $\Omega$ is bounded by the unit circle, and so I don't see why it obey the requirements of the Green's.
Thanks in advance!
You say "$\Omega$ is bounded by the unit circle" but this is too vague/not true.
The curve you want to use is the unit circle centered at the origin, that's right.
The region that this curve bounds is the unit disc.
But the vector field is not even defined in the whole unit disc, so Green's Theorem does not apply.