I need to change $\omega = dx \wedge dy$ to polar coordinate. I have
\begin{align} \omega &= dx \wedge dy\\ &=d(r \cos \theta)\wedge d(r \sin \theta)\\ &=(\cos \theta d r-r \sin \theta d \theta)\wedge (\sin \theta dr + r \cos \theta d \theta)\\ &=\cos \theta \sin \theta \, dr \wedge dr+r \cos^2 \theta\, dr \wedge d \theta-r \sin^2 \theta d \theta \wedge dr - r^2 \sin \theta \cos \theta \, d \theta \wedge d \theta\\ &=r \cos^2 \theta\, dr \wedge d \theta + r \sin^2 \theta \, dr \wedge d \theta\\ &= r(\cos^2 \theta+\sin^2 \theta)\, dr \wedge d \theta\\ &= r \, dr \wedge d \theta \end{align}
Is what I have written down the correct manipulation of $\omega$? Here, the $5$th equality holds since $dr \wedge dr = 0$ and as $(dr \wedge d \theta)=-(d \theta \wedge dr)$. Basically wanna know if I am computing the differential forms correctly using the wedge product. This is Exercise $14.19$ of page $361$ out of John M. Lee's Introduction to Smooth Manifolds.