TITLE: How to compute the following Jacobian
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.
The Jacobian determinant of this transformation is therefore
\begin{alignat*}{1}
 & J_n(r, \theta^1, \dots, \theta^{n-1}) \\
 :=& \det\left(\frac{\partial x^i}{\partial r} \bigg| \frac{\partial x^i}{\partial \theta^j}\right)_{\substack{1\leq i\leq n \\ 1\leq j\leq n-1}} \\
 =& \det\begin{pmatrix}
  c_1             & -rs_1            & 0              & \dots & 0              & 0 \\
  s_1c_2           & rc_1c_2           & -rs_1s_2          & \dots & 0              & 0 \\
  s_1s_2c_3          & rc_1s_2c_3         & rs_1c_2c_3         & \dots & 0              & 0 \\
  \vdots           & \vdots           & \vdots           & \ddots & \vdots           & \vdots \\
  s_1s_2\dots s_{n-2}c_{n-1} & rc_1s_2\dots s_{n-2}c_{n-1} & rs_1c_2\dots s_{n-2}c_{n-1} & \dots & rs_1s_2\dots c_{n-2}c_{n-1} & -rs_1s_2\dots s_{n-2}s_{n-1} \\
  s_1s_2\dots s_{n-2}s_{n-1} & rc_1s_2\dots s_{n-2}s_{n-1} & rs_1c_2\dots s_{n-2}s_{n-1} & \dots & rs_1s_2\dots c_{n-2}s_{n-1} & rs_1s_2\dots s_{n-2}c_{n-1}
 \end{pmatrix} \\
 =& c_1\det A + rs_1\det B
\end{alignat*}
(sorry for the layout), where
\begin{alignat*}{1}
 A :=& \begin{pmatrix}
  rc_1c_2           & -rs_1s_2          & \dots & 0              & 0 \\
  rc_1s_2c_3         & rs_1c_2c_3         & \dots & 0              & 0 \\
  \vdots           & \vdots           & \ddots & \vdots           & \vdots \\
  rc_1s_2\dots s_{n-2}c_{n-1} & rs_1c_2\dots s_{n-2}c_{n-1} & \dots & rs_1s_2\dots c_{n-2}c_{n-1} & -rs_1s_2\dots s_{n-2}s_{n-1} \\
  rc_1s_2\dots s_{n-2}s_{n-1} & rs_1c_2\dots s_{n-2}s_{n-1} & \dots & rs_1s_2\dots c_{n-2}s_{n-1} & rs_1s_2\dots s_{n-2}c_{n-1}
 \end{pmatrix}
\end{alignat*}
and
\begin{alignat*}{1}
 B :=& \begin{pmatrix}
  s_1c_2           & -rs_1s_2          & \dots & 0              & 0 \\
  s_1s_2c_3          & rs_1c_2c_3         & \dots & 0              & 0 \\
  \vdots           & \vdots           & \ddots & \vdots           & \vdots \\
  s_1s_2\dots s_{n-2}c_{n-1} & rs_1c_2\dots s_{n-2}c_{n-1} & \dots & rs_1s_2\dots c_{n-2}c_{n-1} & -rs_1s_2\dots s_{n-2}s_{n-1} \\
  s_1s_2\dots s_{n-2}s_{n-1} & rs_1c_2\dots s_{n-2}s_{n-1} & \dots & rs_1s_2\dots c_{n-2}s_{n-1} & rs_1s_2\dots s_{n-2}c_{n-1}
 \end{pmatrix},
\end{alignat*}
where we have developed the determinant by minors along the first row.
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*}
Overall, we obtain that
\begin{alignat*}{1}
 J_n(r, \theta^1, \dots, \theta^{n-1}) &= c_1\det A + rs_1\det B \\
 &= rc_1^2s_1^{n-2}J_{n-1}(r,\theta^2, \dots\theta^{n-1}) + rs_1^nJ_{n-1}(r,\theta^2, \dots\theta^{n-1}) \\
 &= r(1-s_1^2)s_1^{n-2}J_{n-1}(r,\theta^2, \dots\theta^{n-1}) + rs_1^nJ_{n-1}(r,\theta^2, \dots\theta^{n-1}) \\
 &= rs_1^{n-2}J_{n-1}(r,\theta^2, \dots\theta^{n-1}).
\end{alignat*}
We calculate directly that
\begin{alignat*}{1}
 J_2(r, \theta^1) &= r \\
 J_3(r, \theta^1, \theta^{2}) &= r^2\sin\theta^1.
\end{alignat*}
It follows by induction that
\begin{align*}
 J_n(r, \theta^1, \dots, \theta^{n-1}) = r^{n-1}\sin^{n-2}\theta^1\sin^{n-3}\theta^2\dots\sin\theta^{n-2}.
\end{align*}