On which open set in $\mathbb{R}^2$ is the $1$-form $(xdy-ydx)/(x^2+y^2)$ exact

Question

I think on an open set in the first quadrant not containing zero, there exists a function $f$ such that $df(x,y) = (xdy-ydx)/(x^2+y^2)$.

I think its like solving a differential equation with $x=rcos \theta$ and $y=rsin \theta$.

But I am not being able to do anything rigorous. Kindly help.

Answer 1

Yes, it is exact differential, see $$\frac{d \tan^{-1} (y/x)}{dx}=\frac{1}{1+y^2/x^2} \frac{xdy-ydx}{x^2}=\frac{xdy-ydx}{x^2+y^2}, (x,y) \ne (0, 0)$$