Potential function of $\textbf{F}\left ( x, y \right )=\left ( -\frac y{x^2+y^2},\frac x{x^2+y^2} \right )$ Let $\textbf{F}\left ( x, y \right )=\left ( -\frac y{x^2+y^2},\frac x{x^2+y^2} \right )$ be a vector field in $\mathbb{R}^2-\left \{ \textbf{0} \right \}$.
I know that the potential function of $\textbf{F}$ on $x>0$ is $\arctan \left ( \frac yx \right )$.
But I want to know the potential function of $\textbf{F}$ on $\mathbb{R}^2-\left \{ \textbf{0} \right \}$. Does it exist?
 A: Nope, unfortunately it doesn't, and the way we know this is by studying whether or not the set we're dealing with (in this case $\mathbb R^2 \smallsetminus\{0\}$) is simply connected. Simply connected in layman's terms means "sufficiently un-hole-y" if that makes sense... Maybe "without too many holes" is better. This is obviously a bit far from being a rigorous mathematical statement, so I'll go into more detail: $$\text{A subset $U$ of $\mathbb R^n$ is said to be $\textit{simply connected}$ if any closed curve can be continuously} \\\text{ transformed to a point.}$$
This is still somewhat informal, but it highlights the meaning a bit more. We want to be able to contract a circle to a point without passing over holes. A few examples: $$\left\{\begin{align}&\text{The punctured plane, $\mathbb R^2\smallsetminus\{0\}$, is not simply connected.} \\ &\text{The plane minus a line, $\mathbb R^2\smallsetminus \mathrm {span}\left(u\right)$, is simply connected.}\\ &\text{$\mathbb R^3\smallsetminus{\mathrm {span}(v)}$ is not simply connected.} \\ &\text{$\mathbb R^3\smallsetminus\{0\}$ is simply connected.} \\&\text{The 2-sphere is simply connected.} \\ &\text{A torus is not simply connected.}\end{align}\right.$$
Do you see why these statements are true (intuitively)?
Among other equivalences, a vector field $F:U\subset\mathbb R^2\to\mathbb R^2$ is conservative $\equiv U$ is simply connected and $\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}=0 $. In your example the second condition holds but not the first, so $F$ cannot be conservative!
Thus far I am sure of, though the topic expands leagues beyond (which I haven't yet studied much), namely the question of, when $\mathrm d \omega=0$, what topological conditions (such as simple-connectedness) allow us to say that $\omega$ is an exact form, which is sort of like saying "is-the-derivative-of" something. Think of the $\mathrm d$ as a kind of generalized derivative, or google differential forms. Also try to identify these terms with the problem we're dealing with here!
I don't know much else, but I know this area of study is called De-Rham cohomology, and I find it fascinating. So your question may be deeper than you think!
A: If $\mathbf{F}=\nabla\Phi$, then the integral around the path $(x,y)=(\cos(t),\sin(t))$ for $t\in[0,2\pi]$,
$$
\begin{align}
\oint\mathbf{F}(x,y)\cdot\mathrm{d}(x,y)
&=\int_0^{2\pi}(-\sin(t),\cos(t))\cdot(-\sin(t),\cos(t))\,\mathrm{d}t\\
&=\int_0^{2\pi}1\,\mathrm{d}t\\[4pt]
&=2\pi
\end{align}
$$
should be $\Phi(1,0)-\Phi(1,0)=0$.
