Show that in changing from rectangular coordinates $(x,y)$ to polar coordinates $(r,\theta )$, the Jacobian of $x$ and $y$ with respect to $r$ and $\theta$ is $r$.

Solution To change the variables from rectangular to polar coordinates, we use the equations $$\begin{equation*} x=r\cos \theta \qquad y=r\sin \theta \end{equation*}$$

The partial derivatives are $$\dfrac{\partial x}{\partial r}=\cos \theta \qquad \dfrac{\partial x}{ \partial \theta }=-r\sin \theta \qquad \dfrac{\partial y}{\partial r}=\sin \theta \qquad \dfrac{\partial y}{\partial \theta }=r\cos \theta$$

The Jacobian of $x,y$ with respect to $r,\theta$ is $$\begin{eqnarray*} \dfrac{\partial (x,y)}{\partial (r,\theta )}&=&\left\vert \begin{array}{c@{\quad}c} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \theta } \\[9pt] \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \theta } \end{array} \right\vert =\left\vert \begin{array}{l@{\quad}r} \cos \theta & -r\sin \theta \\[3pt] \sin \theta & r\cos \theta \end{array} \right\vert =\cos \theta \cdot r\cos \theta -\sin \theta (-r\sin \theta)\\[4pt] &=&r\cos ^{2}\theta +r\sin ^{2}\theta =r \end{eqnarray*}$$