I'm trying to convert formally the cartesian differential product $dxdy$ into its equivalent in the cylindrical coordinates which is $rdrd \theta$ . Doing the change of variable: $x=rcos\theta$ and $y=rsin\theta$.I got:
$dx=cos\theta dr-rsin\theta d \theta$
$dy=sin\theta dr+rcos\theta d\theta$
However the usual product $\times$ between $dx$ and $dy$ doesn't give $rdrd \theta$. Any hints?
As Sean Roberson said in the comments, you'll need to compute the determinant of the Jacobian since you are trying to rewrite the differential area and the Jacobian of the coordinate conversion formula has to be considered. Specifically, the relation is given as follows: $$ dxdy= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\ \end{vmatrix}dr\;d\theta$$ It seems you have most of these calculated already. Formally, if you let $f:U\to\mathbb{R}^2$, where $U\subseteq\mathbb{R}^2$ consists of all pairs $(r,\theta)$ with $r>0, \theta \in\mathbb{R}$. Let $f(r,\theta)=(r\cos\theta,r\sin\theta)$, then the Jacobian of $f$ is: $$J_f(r,\theta)=\begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \\ \end{pmatrix}$$ Calculating the determinant: $ \textrm{Det}J_f(r,\theta)=r\cos^2\theta+r\sin^2\theta=r$, and we readily arrive at $dxdy=rdrd\theta$.
