Why isn't $dydx$ equal to $r\cos^2\theta dr\,d\theta$?

Question

$y = r\sin(\theta)$, $x = r\cos\theta$

$$dy = r\cos\theta d\theta$$

$$dx = \cos \theta dr$$

$$dydx = r\cos^2\theta dr d\theta$$

What did I do wrong? Also I don't know how to format sorry

Answer 1

You need to take total differentials $$ dx=\cos\theta dr-r\sin\theta d\theta;\quad dy=\sin\theta dr+r\cos\theta d\theta $$ and then you take their exterior product (when you write $dxdy$ what you actually mean is $dx\wedge dy$) $$ dx\wedge dy=(\cos\theta dr-r\sin\theta d\theta)\wedge(\sin\theta dr+r\cos\theta d\theta)=r\cos^2\theta dr\wedge d\theta-r\sin^2\theta d\theta\wedge dr= $$ $$ =r\cos^2\theta dr\wedge d\theta+r\sin^2\theta dr\wedge d\theta=rdr\wedge d\theta. $$