I have a vector field $\;\underline{v}(x,y) = \left(\dfrac{-y}{x^{2}+4y^{2}}, \dfrac{x}{x^{2}+4y^{2}}\right)$
$\text{curl }\underline{v} = 0$ and thus the vector field is conservative.
I have a surface, $\Omega$, with $\Omega:=\left\{(x,y)\in \mathbf{R}^2: x^2+4y^2=4\right\}$.
A parameterisation of the border of $\Omega$, $\partial\Omega$ is $\underline{r}(t) = \big(2\cos(t), \sin(t)\big)$, for $t \in [0, 2\pi]$
The integral $\oint_{\partial\Omega}\underline{v}\cdot d\underline{r} = \pi$
However, by Green's Theorem, $\oint_{\partial\Omega}\underline{v}\cdot d\underline{r} = \iint_{\Omega}\textbf{curl}\,\underline{v}\text{ }dxdy$. Since curl $\underline{v} = 0$, and the curve is closed, $\oint_{\partial\Omega}\underline{v}\cdot d\underline{r} = 0$
Why are these two results not contradictory?
Edit: I think that this might be because $\underline{v}(0,0)$ is undefined, however I am not sure what difference that would make, nor how to formalise why this is not a contradiction.
By definition, a conservative vector field is one which is the gradient of some scalar function. In $\mathbb{R}^2$, vector fields which have zero curl are only conservative on domains without holes (for example, simply connected regions). Since your vector field is only defined on $\mathbb{R}^2\setminus\{0\}$ (which has a hole) it is not conservative.
However, if you already know the vector field is conservative, it's always true that its curl is zero. This follows from the fact that $\nabla\times(\nabla \phi)=0$.