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本帖最后由 hbghlyj 于 2022-3-18 15:45 编辑 $V_n(R)$ and $S_n(R)$ are the $n$-dimensional volume of the $n$-ball and the surface area of the $n$-sphere embedded in dimension $n+1$, respectively, of radius $R$.
The constants $V_n$ and $S_n$ (for $R=1$, the unit ball and sphere) are related by the recurrences:
$$\begin{aligned}V_{0}&=1&V_{n+1}&={\frac {S_{n}}{n+1}}\\[6pt]S_{0}&=2&S_{n+1}&=2\pi V_{n}\end{aligned}$$
The surface area, or properly the n-dimensional volume, of the n-sphere at the boundary of the (n + 1)-ball of radius R is related to the volume of the ball by the differential equation$$S_{n}R^{n}={\frac {dV_{n+1}R^{n+1}}{dR}}={(n+1)V_{n+1}R^{n}},$$
or, equivalently, representing the unit $n$-ball as a union of concentric $(n − 1)$-sphere shells,
$$V_{n+1}=\int _{0}^{1}S_{n}r^{n}\,dr.$$
So,
$$V_{n+1}={\frac {S_{n}}{n+1}}.$$
We can also represent the unit $(n+2)$-sphere as a union of products of a circle (1-sphere) with an $n$-sphere. Let $r=\cos θ$ and $r^2+R^2=1$, so that $R =\sin θ$ and $dR =\cos θ dθ$. Then,
$$\begin{aligned}S_{n+2}&=\int _{0}^{\frac {\pi }{2}}S_{1}r\cdot S_{n}R^{n}\,d\theta \\[6pt]&=\int _{0}^{\frac {\pi }{2}}S_{1}\cdot S_{n}R^{n}\cos \theta \,d\theta \\[6pt]&=\int _{0}^{1}S_{1}\cdot S_{n}R^{n}\,dR\\[6pt]&=S_{1}\int _{0}^{1}S_{n}R^{n}\,dR\\[6pt]&=2\pi V_{n+1}.\end{aligned}$$
Since $S_1 = 2π V_0$, the equation
$$S_{n+1}=2\pi V_{n}.$$
holds for all $n$.
In general, the volume, in n-dimensional Euclidean space, of the unit $n$-ball, is given by
$$V_{n}={\frac {\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}+1\right)}}={\frac {\pi ^{\frac {n}{2}}}{\left({\frac {n}{2}}\right)!}}$$By multiplying $V_n$ by $R_n$, differentiating with respect to $R$, and then setting $R=1$, we get the closed form
${\displaystyle S_{n-1}={\frac {n\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}+1\right)}}={\frac {2\pi ^{\frac {n}{2}}}{\Gamma \left({\frac {n}{2}}\right)}}.}$
for the $(n − 1)$-dimensional volume of the sphere $S^{n−1}$. |
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