Question

The formulas for converting from cylindrical coordinates to rectangular coordinates are x = r cos θ, y = r sin θ, and z = z. Prove that the Jacobian $\frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r$.

Short answer

It is proven that$\frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r$

Step-by-step solution

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Given information

The transformation equations are,

$x=r\cos \theta ,y=r\sin \theta ,z=z$

Proof

The definition of Jacobian of a transformation using partial derivatives is given as

$$\begin{gathered} L.H.S=\frac{\partial (x,y,z)}{\partial (r,\theta ,z)} \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=\left|\begin{array}{ccc}\frac{\partial x}{\partial r}& \frac{\partial y}{\partial r}& \frac{\partial z}{\partial r}\\ \frac{\partial x}{\partial \theta }& \frac{\partial y}{\partial \theta }& \frac{\partial z}{\partial \theta }\\ \frac{\partial x}{\partial z}& \frac{\partial y}{\partial z}& \frac{\partial z}{\partial z}\end{array}\right| \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=\left|\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -r\sin \theta & r\cos \theta & 0\\ 0& 0& 1\end{array}\right| \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r{\cos }^2\theta +r{\sin }^2\theta \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r({\cos }^2\theta +{\sin }^2\theta ) \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=r \\ \frac{\partial (x,y,z)}{\partial (r,\theta ,z)}=R.H.S \end{gathered}$$

Hence, proved.