# How to compute the following Jacobian

I need to show that the Jacobian of the n-dimensional spherical coordinates is $$\displaystyle r^{n-1}\sin^{n-2}\phi_1\sin^{n-3}\phi_2\cdots\sin\phi_{n-2}$$ then I have computed the Jacobian matrix, and is consisten with this one http://faculty.madisoncollege.edu/alehnen/sphere/Apendxa/Appendixa.htm But the thing is, how to compute the determinant of that huge matrix.

Thanks a lot for your help.

• Do you know about minors? Commented Apr 15, 2015 at 14:30
• Yes of course :) Commented Apr 15, 2015 at 14:30
• I haven't tried, but I'm guessing induction would be the way to go. Commented Apr 15, 2015 at 14:30
• Right but I think it is so hard to guess the steps, I was considering but is difficult for me :( Commented Apr 15, 2015 at 14:32
• Well I think the should be a better way, or how can you proceed by induction? Commented Apr 15, 2015 at 14:34

General spherical coordinates in $\mathbb{R}^n$ are given by \begin{alignat*}{1} x^1 &= r\cos\theta^1 \\ x^2 &= r\sin\theta^1\cos\theta^2 \\ x^3 &= r\sin\theta^1\sin\theta^2\cos\theta^3 \\ &\vdots \\ x^{n-1} &= r\sin\theta^1\sin\theta^2\dots\sin\theta^{n-2}\cos\theta^{n-1} \\ x^n &= r\sin\theta^1\sin\theta^2\dots\sin\theta^{n-2}\sin\theta^{n-1}. \end{alignat*} We note that \begin{align*} \frac{\partial x^1}{\partial r} = \cos\theta^1 = c_1, \quad \frac{\partial x^1}{\partial \theta^1} = -r\sin\theta^1 = -rs_1, \quad \frac{\partial x^1}{\partial \theta^j} = 0\text{ for }j>1, \end{align*} where we have set \begin{align*} c_j := \cos\theta^j\text{ and } s_j := \sin\theta^j \end{align*} for brevity.
Let us consider $\det A$. We pull the factor $rc_1$ out of the first column of $A$ and $s_1$ out of the $n-2$ other columns, so \begin{alignat*}{1} \det A &= rc_1s_1^{n-2}\det\begin{pmatrix} c_2 & -rs_2 & \dots & 0 & 0 \\ s_2c_3 & rc_2c_3 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ s_2\dots s_{n-2}c_{n-1} & rc_2\dots s_{n-2}c_{n-1} & \dots & rs_2\dots c_{n-2}c_{n-1} & -rs_2\dots s_{n-2}s_{n-1} \\ s_2\dots s_{n-2}s_{n-1} & rc_2\dots s_{n-2}s_{n-1} & \dots & rs_2\dots c_{n-2}s_{n-1} & rs_2\dots s_{n-2}c_{n-1} \end{pmatrix} \\ &= rc_1s_1^{n-2}J_{n-1}(r,\theta^2, \dots\theta^{n-1}) \end{alignat*}
Similarly, to calculate $\det B$, we can pull $s_1$ out of all $n-1$ columns of $B$: \begin{alignat*}{1} \det B &= s_1^{n-1}\det\begin{pmatrix} c_2 & -rs_2 & \dots & 0 & 0 \\ s_2c_3 & rc_2c_3 & \dots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ s_2\dots s_{n-2}c_{n-1} & rc_2\dots s_{n-2}c_{n-1} & \dots & rs_2\dots c_{n-2}c_{n-1} & -rs_2\dots s_{n-2}s_{n-1} \\ s_2\dots s_{n-2}s_{n-1} & rc_2\dots s_{n-2}s_{n-1} & \dots & rs_2\dots c_{n-2}s_{n-1} & rs_2\dots s_{n-2}c_{n-1} \end{pmatrix} \\ &= s_1^{n-1}J_{n-1}(r,\theta^2, \dots\theta^{n-1}). \end{alignat*}