From the theorem of open simply-connected regions we know that if $\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}$ then $\vec F(x,y)=P\vec i+Q\vec j$ is conservative, well I have two examples related to this that I'm not understanding.
Example 1:
$\vec F(x,y)=\left(\frac{x}{(x^2+y^2)^{3/2}},\frac{y}{(x^2+y^2)^{3/2}}\right)$
$\text{Domain of $\vec F(x,y)$:} \mathbb{R}^2-\{(0,0)\}$
$\text{potential:} f(x,y)=\frac{-1}{\sqrt{x^2+y^2}}$
$\text{Domain of $\vec f(x,y)$:} \mathbb{R}^2-\{(0,0)\}$
Example 2:
$\vec F(x,y)=\left(-\frac{y}{(4x^2+y^2)},\frac{x}{(4x^2+y^2)}\right)$
$\text{Domain of $\vec F(x,y)$:} \mathbb{R}^2-\{(0,0)\}$
$\text{potential:} f(x,y)=\frac{\tan^{-1}\left(\frac{y}{2x}\right)}{2}$
$\text{Domain of $\vec f(x,y)$:} \{(x,y), x\neq 0\}$
The argument for the example 1 one being a conservative field it's because the function of the field has the same domain of the potential function, and the argument for the second example not being a conservative field it's because the domain of the potential function is greater than the domain of the field and because it's line integral along a closed curve isn't zero, but if we didn't used this arguments, based on the theorem above why wouldn't both fields not be conservative, because they have restrictions on their domain so they have a "hole" on the region and wouldn't they fail to be open and simply-connected?
Best Answer
The second example is essentially the same as here. In short: it is not conservative because the potential in polar coordinates is the polar angle which has a discontinuity at the positive $x$-axis by which a integral over a closed loop is not zero.
The most rigorous way to show conservativity of the first example is I think the following: It is the restriction to the $xy$-plane of the vector field $$\tag{1} \vec{F}=\frac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}} $$ which has potential $f=\frac{-1}{\sqrt{x^2+y^2+z^2}}\,.$ Both are defined on the simply connected domain $\mathbb R^3\setminus\{0\}\,.$ (Note that $\mathbb R^\color{red}{2}\setminus\{0\}$ is not simply connected.) The conservativity of (1) in 3d carries over trivially to the two-dimensional case because we just have to consider loops in the $xy$-plane.
Another way of looking at it is this: Every closed loop is the boundary of a surface that does not contain the origin and the integral over that surface is zero because of ${\rm curl}\,\vec{F}=0\,.$ Hence, by Stokes' theorem $\oint \vec{F}\cdot \,d\vec{\gamma}=0\,.$ In particular this holds for loops in the $xy$-plane.