Finding a Jacobian

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*}