Differential form calculation

Question

Below is a problem from Arnold's Mathematical Methods of Classical Mechanics. I'm not seeing how the calculation for $\omega_3$ is performed. Any help would be appreciated.

Answer 1

I will denote $x_{1}=x$ and $x_{2}=y$.

Using the identities $x=r\cos\phi$ and $y=r\sin\phi$ you have $r=\sqrt{x^2+y^2}$ and $\phi=\tan^{-1}(\frac{y}{x})$.

Now compute $dr=\partial_{x}rdx+\partial_{y}rdy$ and $d\phi=\partial_{x}\phi dx+\partial_{y}\phi dy$.

You will obtain $\omega_{3}(x,y) =dx\wedge dy$.