What is the Jacobian in this transformation

Question

I have a transformation \begin{equation} x = r\left(\frac{v_y\sin\theta\cos\phi+v_xv_z\sin\theta\sin\phi}{\sqrt{1-v_z^2}}+v_x\cos\theta\right)~. \end{equation} \begin{equation} y = r\left(\frac{-v_x\sin\theta\cos\phi+v_yv_z\sin\theta\sin\phi}{\sqrt{1-v_z^2}}+v_y\cos\theta\right)~. \end{equation} \begin{equation} z = r(v_z\cos\theta-\sqrt{1-v_z^2}\sin\theta\sin\phi)~, \end{equation} I would like to know how the following transformation would look like: $$\mathrm{d}x\mathrm{d}y\mathrm{d}z\rightarrow (r, \theta, \phi)~.$$ I know I should use the Jacobian matrix in this case: \begin{equation} J= \begin{pmatrix} \frac{\mathrm{d}x}{\mathrm{dr}} & \frac{\mathrm{d}x}{\mathrm{d}\theta} & \frac{\mathrm{d}x}{\mathrm{d}\phi}\\ \frac{\mathrm{d}y}{\mathrm{dr}} & \frac{\mathrm{d}y}{\mathrm{d}\theta} & \frac{\mathrm{d}y}{\mathrm{d}\phi}\\ \frac{\mathrm{d}z}{\mathrm{dr}} & \frac{\mathrm{d}z}{\mathrm{d}\theta} & \frac{\mathrm{d}z}{\mathrm{d}\phi}~. \end{pmatrix}~, \end{equation} and find its determinant. But, I couldn't get a reasonable simplification. Does anyone know if this actually has some simple expression, or its quite complicated?

Answer 1

I find for the determinant this simple expression

$$d = \frac{r^2 \sin (\theta ) \left(v_{x}^2+v_{y}^2\right)}{1-v_{z}^2}$$