Prove $dx*dy = r*dr*dφ$ using $d(r*cosφ)*d(r*sin(φ))$

Question

I am trying to demonstrate that $dx*dy$ (in cartesian coordinates) is equal to $r*dr*dφ$ (polar coordinates). I know the image, but I want to follow an other way:

$$x=r*cosφ$$ $$y=r*sinφ$$

$$d(r*cosφ)*d(r*sin(φ)) \rightarrow ?$$

Answer 1

I suggest you use Jacobian matrix since this is not very rigorous. $$\begin{align*} |d(r \cos \phi)| |d(r \sin \phi)| &= |dr \cos \phi + r\,d\cos\phi| |dr \sin \phi + r\,d\sin \phi| \\ &= |\cos \phi\,dr - r \sin \phi\,d\phi| |\sin\phi\,dr + r \cos\phi\,d\phi| \\ &= r(\sin\phi)^2\,dr\,d\phi + r(\cos\phi)^2\,dr\,d\phi \end{align*}$$ The $(dr)^2$ terms and $(d\phi)^2$ terms are negligible.

Answer 2

In the $x,y$ co-ordinate system element of area or differential of area is $ dx\cdot dy $ (red rectangle).

In the $r,\varphi$ co-ordinate system element of area or differential of area is $ dr\cdot r d \varphi $ ( gray sector segment ).

The infinitesimal differential areas ( or volumes) can be computed treating them as if the sides are finite. It is directly geometrical.

The same result is obtained using Jacobian conversion. Recently I also read that Leibnitz used such product evaluations when Newton treated more using time as differential.I used to think it was some sort of engineering shortcut!

$$ dx\cdot dy = dr \cdot r d \varphi $$