I need to solve this problem for my vector calculus class
Let $\mathbf{F}=\frac{-y}{x^2+y^2}\mathbf{i}+\frac{x}{x^2+y^2}\mathbf{j}$, show that $\mathbf{F}$ is a conservative vector field in $\Omega_{1}=\{-\infty<x<\infty,y>0\}\subset\mathbb{R}^2$, but it is not a conservative vector field in $\Omega_{2}=\{0<x^2+y^2<4\}\subset\mathbb{R}^2$.
MY ANSWER
To answer that question, I used the fact that a vector field is conservative if $\text{ }\frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y}=0$.
So I thought I would get something that make the vector field conservative when $y>0$.
But here's what I get: $$\begin{eqnarray} \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right)-\frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right)&=&\frac{1}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2}-\left(\frac{2y^2}{(x^2+y^2)^2}-\frac{1}{x^2+y^2}\right)\\ &=&\frac{2}{x^2+y^2}-\frac{2x^2}{(x^2+y^2)^2}-\frac{2y^2}{(x^2+y^2)^2}\\ &=&\frac{2(x^2+y^2)-2x^2-2y^2}{(x^2+y^2)^2}\\ &=&\frac{2x^2+2y^2-2x^2-2y^2}{(x^2+y^2)^2}=0 \end{eqnarray}$$
Therefore, according to what I just found, no matter what the value of $x$ and $y$ is, the vector field will always be conservative. But I am supposed to prove that it is not when $0<x^2+y^2<4$. Where did I make my mistake?
Thanks in advance for your answers.
One way to think about this is that the two-dimensional curl that you calculated isn't always $0$. Specifically, $\frac{0}{(x^2+y^2)^2}$ is undefined when the denominator is zero. In some sense all of the curl is concentrated at that single point.
It may help to visualize this vector field https://www.desmos.com/calculator/jqa3gpyas4 when analyzing if it's conservative.