TITLE: Proving $dx.dy = r.drd\theta$ using x and y
QUESTION [6 upvotes]: I can prove that $dx dy = r.dr d\theta$ by drawing a circle and calculating the area of a small square in the polar coordinates, but when I try proving it using the equations below, I fail to prove it. What is my mistake?
$x = r\cos(\theta) => dx = \cos(\theta).dr - \sin(\theta)r.d\theta$
$y = r\sin(\theta)=>dy=\sin(\theta).dr + \cos(\theta)r.d\theta$
$=> dxdy = r(\cos^2(\theta)-\sin^2(\theta))drd\theta=r\cos(2\theta).drd\theta$

I actually saw a similar question in this site: how to get $dx\; dy=r\;dr\;d\theta$
My problem was that I didn't understand why $drd\theta = -d\theta dr$

REPLY [5 votes]: If you want to do things algebraically then you'll have to resort to the exterior product:
$$\Bbb d x \wedge \Bbb d y = (\cos \theta \ \Bbb d r - r \sin \theta \ \Bbb d \theta) \wedge (\sin \theta \ \Bbb d r + r \cos \theta \ \Bbb d \theta) = \\
\cos \theta \sin \theta \ \Bbb d r \wedge \Bbb d r + r \cos^2 \theta \ \Bbb d r \wedge \Bbb d \theta - r \sin^2 \theta \ \Bbb d \theta \wedge \Bbb d r - r^2 \sin \theta \cos \theta \ \Bbb d \theta \wedge \Bbb d \theta$$
and, since $\Bbb d r \wedge \Bbb d r = \Bbb d \theta \wedge \Bbb d \theta = 0$ and $\Bbb d r \wedge \Bbb d \theta = - \Bbb d \theta \wedge \Bbb d r$ because the exterior product is graded-commutative (in particular anti-commutative for $1$-forms), we get
$$\Bbb d x \wedge \Bbb d y = r \ \Bbb d r \wedge \Bbb d \theta .$$