Spherical Coordinates
The conversion between the 2-d Cartesian coordinate system and the 2-d polar coordinate system can be extended to a higher dimension, say \(k\)-d. In \(k\)-d case, their conversion can be described as below:
Spherical to Cartesian \[ \begin{alignat}{2} r &= \sqrt{x_1^2 + \dots + x_k^2} && \\ \varphi_1 &= \arccot \frac{x_1} {\sqrt{x_k^2 + \dots + x_2^2}} &&= \arccos \frac{x_1} {\sqrt{x_k^2 + \dots + x_1^2}} \\ \varphi_2 &= \arccot \frac{x_2} {\sqrt{x_k^2 + \dots + x_3^2}} &&= \arccos \frac{x_2} {\sqrt{x_k^2 + \dots + x_2^2}} \\ & \vdots &&\vdots\\ \varphi_{k-2} &= \arccot \frac{x_{k-1}} {\sqrt{x_k^2 + x_{k-1}^2}} &&= \arccos \frac{x_{k-2}} {\sqrt{x_k^2 + x_{k-1}^2 + x_{k-2}^2}} \\ \varphi_{k-1} &= 2 \arccot \frac{x_{k-1} + \sqrt{x_k^2 + x_{k-1}^2}}{x_k} &&= \begin{cases} \arccos \frac{x_{k-1}} {\sqrt{x_k^2 + x_{k-1}^2}}, &x_n \ge 0 \\ 2\pi - \arccos \frac{x_{k-1}} {\sqrt{x_k^2 + x_{k-1}^2}}, &x_n > 0\\ \end{cases} \end{alignat} \]
Cartesian to spherical \[ \begin{align} x_1 &= r \cos(\varphi_1) \\ x_2 &= r \sin(\varphi_1) \cos(\varphi_2) \\ \notag &\vdots \\ x_{k-1} &= r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \cos(\varphi_{k-1}) \\ x_k &= r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \sin(\varphi_{k-1}) \\ \end{align} \] The corresponding Jacobian Matrix is \[ J^{(k)} = \left[ \begin{array}{ccccc|c} \cos (\varphi_1) & -r \sin(\varphi_1) & 0 & 0 & \cdots & 0 \\ \sin(\varphi_1) \cos(\varphi_2) & r \cos(\varphi_1) \cos(\varphi_2) & -r \sin(\varphi_1) \sin(\varphi_2) & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ & \ddots & \vdots \\ \hline \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \cos(\varphi_{k-1}) & \cdots & \cdots & \ & \ & -r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \sin(\varphi_{k-1}) \\ \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \sin(\varphi_{k-1}) & \cdots & \cdots & \ & \ & r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \cos(\varphi_{k-1}) \end{array} \right] \] \(|J^{(2)}|\) can be easily derived as \(r\); and \(|J^{(k)}|\) can be constructed from \(|J^{(k-1)}|\). Comparing \(J^{(k)}\) and \(J^{(k-1)}\),
- \(J^{(k)}\) has an extra column \(k\) and an extra row \(k\),
- On row \(k-1\), before column \(k\), \(J^{(k)}\) has an extra \(\cos(\varphi_{k-1})\) term in each element than the elements of \(J^{(k-1)}\) on row \(k-1\),
- On row \(k\), before column \(k\), \(J^{(k)}\) has an extra \(\sin(\varphi_{k-1})\) term in each element than the elements of \(J^{(k-1)}\) on row \(k-1\),
- \(J^{(k)}\) and \(J^{(k-1)}\) are totally the same on the region delimited by row \(1\), row \(k-2\), column \(1\), column \(k-1\).
Apply the Laplace expansion along column \(k\) and combine the property of determinant to give \[ \begin{aligned} |J^{(k)}| =\ & \underbrace{0 + \dots + 0}_{n-2} + (-1)^{(n-1)+n} [-r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \sin(\varphi_{k-1}) \sin(\varphi_{k-1}) \big( \sin(\varphi_{k-1}) |J^{(k-1)}| \big)] + \\ & (-1)^{n+n} [r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \cos(\varphi_{k-1}) \big( \cos(\varphi_{k-1}) |J^{(k-1)}| \big)] \\ =\ & r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) \big( \sin_{\varphi_{k-1}}^2 + \cos{\varphi_{k-1}}^2 \big) |J^{(k-1)}| \\ =\ & r \sin(\varphi_1) \dots \sin(\varphi_{k-2}) |J^{(k-1)}| \\ \end{aligned} \] By induction, \[ |J^{(k)}| = r^{k-1} \sin^{k-2}(\varphi_1) \sin^{k-3}(\varphi_2) \dots \sin(\varphi_{k-2}) \] Therefore when changing basis from orthogonal coordinate system to polar coordinate system, \[ \d x_1 \dots \d x_k = r^{k-1} \sin^{k-2}(\varphi_1) \sin^{k-3} (\varphi_2) \dots \sin(\varphi_{k-2}) \d r \d \varphi_1 \dots \d \varphi_{k-1} \]
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Volume of Sphere
With change of basis between spherical coordinate and Cartesian, we may compute the volume of sphere in any \(n\)-dimension. \[ \begin{aligned} V_n &= \int_{B_n} 1 \; \d x_1 \d x_2 \dots \d x_n \\ &= \int_{0}^{R} \int_{0}^{2 \pi} \underbrace{\int_{0}^{\pi} \dots \int_{0}^{\pi}}_{n-2} \\ &\quad r^{n-1} \sin(\varphi_1)^{n-2} \sin(\varphi_2)^{n-3} \dots \sin(\varphi_{n-2}) \d r \d \varphi_1 \dots \d \varphi_{n-1} \\ &= \int_{0}^{R} r^{n-1} \d r \int_{0}^{2 \pi} \d \varphi_{n-1} \\ &\quad \int_{0}^{\pi} \sin(\varphi_{n-2}) \d \varphi_{n-2} \int_{0}^{\pi} \sin^2(\varphi_{n-3}) \d \varphi_{n-3} \dots \int_{0}^{\pi} \sin^{n-2}(\varphi_{1}) \d \varphi_{1} \\ \end{aligned} \] Notice that \[ \int_{0}^{\pi} \sin^{n}(x) \d x = \sqrt{\pi} \frac{\Gamma (\frac{n-1}{2})}{\Gamma (\frac{n}{2})} \] Therefore, \[ \begin{aligned} V_n &= \int_{0}^{R} r^{n-1} \d r \int_{0}^{2 \pi} \d \varphi_{n-1} \\ &\quad \int_{0}^{\pi} \sin(\varphi_{n-2}) \d \varphi_{n-2} \int_{0}^{\pi} \sin^2(\varphi_{n-3}) \d \varphi_{n-3} \dots \int_{0}^{\pi} \sin^{n-2}(\varphi_{1}) \d \varphi_{1} \\ &= \frac{R^n}{n} \cdot 2\pi \cdot \sqrt{\pi} \frac{\Gamma(0)}{\Gamma(1/2)} \cdot \sqrt{\pi} \frac{\Gamma(1/2)}{\Gamma(2/2)} \cdots \sqrt{\pi} \frac{\Gamma((n-3)/2)}{\Gamma((n-2)/2)} \\ &= R^n \frac{1}{n/2} \sqrt{\pi^n} \frac{1}{\Gamma((n-2)/2)} \\ &= \frac{R^n \sqrt{\pi^n}}{\Gamma(n/2)} \end{aligned} \] 《三体》中的数学——为什么很高维度的单位球体积为0？_哔哩哔哩_bilibili