Prove $dx*dy = r*dr*dφ$ using $d(r*cosφ)*d(r*sin(φ))$ 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:
$$x=r*cosφ$$
$$y=r*sinφ$$
$$d(r*cosφ)*d(r*sin(φ)) \rightarrow ?$$
 A: 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.
A: 
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!
$$ dx\cdot dy = dr \cdot  r d \varphi $$ 
