Consider the differential form:
$$\omega=-\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy$$
I believe that this form is not exact. i.e: There does not exist a $\psi$ such that $d\psi=\omega$.
In particular, this is supposed to be true since the line integral along a loop, $$\int_{C}\omega = 2\pi \neq 0$$
But, what about $$\psi=-\arctan(\frac x{y})$$
According to Maple,
As you can see, Maple says that $d\psi=\omega$. So what's the deal here? Is $\omega$ exact or not?
Thank You!
On
Your $\psi$ is just defined for $y\neq 0$. Therefore $d\psi=\omega$ holds on $A=\{(x,y)\in\mathbb R^2~:~ y>0\}$ and on $B=\{(x,y)\in\mathbb R^2~:~y<0\}$. You can conclude that for any closed curve $\gamma$ in $A$ or in $B$ you get $\int_\gamma \omega =0$. But your loop doesn't belong to $A$ or $B$. Moreover, you can conclude from $\int_C\omega\neq 0$ for a loop $C$ that there isn't any $\psi$ such that $d\psi=\omega$ on $\mathbb R^2\setminus\{(0,0)\}$.
The differential form $\omega$ is exact in some region where $\arctan(y/x)$, or more generally the phase $\operatorname{atan}(x,y)$, makes sense. For example, it is exact in the domain you get when you throw away the non-positive half of the $x$-axes.
However, it is not exact in the punctured plane $\Bbb{R}^2\setminus\{0,0\}$. That non-vanishing path integral around the closed loop proves that.