Find the volume of $\mathbb S^n$ (the surface area of $(n+1)$-dimensional unit ball) by using differential geometry

Question

I am reading the first 16 chapters of Lee's Introduction to Smooth Manifolds and trying to find the volume of $\mathbb S^n$.

I know the Riemannian volume form of $\mathbb S^n$ is $$\omega=\sum_{i=1}^{n+1}(-1)^{i-1}x^idx^1\wedge\cdots\wedge \widehat{dx^i}\wedge\cdots\wedge dx^{n+1},$$ and I also know the spherical coordinate systems of $\mathbb R^{n+1}$ is $$F:\overline D\rightarrow\overline{\mathbb B^{n+1}}$$ $$(\rho,\varphi_1,\cdots,\varphi_n)\mapsto\left(\rho\cos\varphi_1,\rho\sin\varphi_1\cos\varphi_2,\rho\sin\varphi_1\sin\varphi_2\cos\varphi_3,\cdots,\rho\left(\prod_{i=1}^{n-1}\sin\varphi_i\right)\cos\varphi_n,\rho\prod_{i=1}^n\sin\varphi_i\right)$$ where $D:=(0,1]\times(0,\pi)^{n-1}\times(0,2\pi)$. But the computation will be very messy and complicated and I'm stuck when I substitute the parameterization in the Riemannian volume form of $\mathbb S^n$. As you can see: \begin{equation*} \begin{aligned} {\rm Vol}(\mathbb S^n) &=\int_{\mathbb S^n}\omega\\ &=\int_D F^*\omega \\ &=\int_D F^*\left(\sum_{i=1}^{n+1}(-1)^{i-1}x^idx^1\wedge\cdots\wedge \widehat{dx^i}\wedge\cdots\wedge dx^{n+1}\right) \\ &=\int_D \sum_{i=1}^n (-1)^{i-1}\left(\rho\left(\prod_{i=1}^{n-1}\sin\varphi_i\right)\cos\varphi_n\right) d(\rho\cos\varphi_1)\wedge\cdots\wedge \widehat{dx^i}\wedge\cdots\wedge d\left(\rho\prod_{i=1}^n\sin\varphi_i\right) \\ &+\int_D (-1)^n \left(\rho\prod_{i=1}^n\sin\varphi_i\right) d(\rho\cos\varphi_1)\wedge\cdots\wedge d\left(\rho\left(\prod_{i=1}^{n-1}\sin\varphi_i\right)\cos\varphi_n\right) \\ &=? \end{aligned} \end{equation*} Any help would be great appreciated!

Answer 1

A computation show that the Jacobian determinant of $F$ is $$\det DF=\rho^n\sin^{n-1}\varphi_1\sin^{n-2}\varphi_2\cdots\sin\varphi_{n-1}>0.$$ So $F$ is orientation-preserving.

Let $\eta:=dx^1\wedge\cdots\wedge dx^{n+1}$ denote the Riemannian volume form of $\mathbb R^{n+1}$ and let $\displaystyle N:=\sum_{i=1}^{n+1}x^i\frac\partial{\partial x^i}$. It's obvious that $N|_{\mathbb S^n}$ is a smooth outward-pointing unit normal vector field along $\mathbb S^n$. Therefore, $$\omega=\iota^*(N\lrcorner~\eta).$$ where $\iota:\mathbb S^n\rightarrow\mathbb R^{n+1}$ is the inclusion map.

Since $\mathbb S^n=\partial\overline{\mathbb B^{n+1}}$, \begin{equation*} \begin{aligned} {\rm Vol}(\mathbb S^n) &=\int_{\mathbb S^n}\omega\\ &=\int_{\mathbb S^n}\iota^*(N\lrcorner~\eta) \\ &=\int_\overline{\mathbb B^{n+1}} ~d(N\lrcorner~\eta)\quad(\text{Stokes's Theorem}) \\ &=\int_\overline{\mathbb B^{n+1}} ~({\rm div} N)~\eta \\ &=\int_\overline{\mathbb B^{n+1}} ~\left(\sum_{i=1}^{n+1}\frac{\partial x^i}{\partial x^i}\right)\eta \\ &=(n+1)\int_\overline{\mathbb B^{n+1}} ~\eta \\ &=(n+1)\int_D F^*\eta \\ &=(n+1)\int_D (\det DF)~d\rho~d\varphi_1~\cdots~d\varphi_n\quad(\text{Proposition 14.20 of Lee's ISM}) \\ &=(n+1)\int_0^1\rho^nd\rho\int_0^\pi\sin^{n-1}\varphi_1d\varphi_1\cdots\int_0^\pi\sin\varphi_{n-1}d\varphi_{n-1}\int_0^{2\pi}d\varphi_n \\ &=\frac{\displaystyle 2\pi^{(n+1)/2}}{\displaystyle \Gamma\left(\frac{n+1}2\right)}. \end{aligned} \end{equation*}

Remark: From the sixth equal sign, we can get $${\rm Vol}(\mathbb S^n)=(n+1){\rm Vol}\left(\overline{\mathbb B^{n+1}}\right).$$ That is, the surface area of $(n+1)$-dimensional unit ball is $n+1$ times its volume.

Answer 2

Here is another solution using a Gaussian integral $$\int_{R^n}e^{-s\|z\|^2/2}dz=\frac{(2\pi)^{n/2}}{s^{n/2}}=C_n\int_0^{\infty}e^{-sx}x^{\frac{n}{2}-1}dx$$ where $C_n=\frac{(2\pi)^{n/2}}{\Gamma(n/2)}.$ This means that the Laplace transform of the image of the Lebesgue measure $dz$ on $R^n$ by the map $z\mapsto x=\|z^2\|/2$ is the Laplace transform of $C_nx^{\frac{n}{2}-1}dx.$ As a consequence the volume of the ball $B(r)$ of radius $r$ is $$V_n(r)=C_n\int_0^{r^2/2}x^{\frac{n}{2}-1}dx=C_n\times \frac{2}{n}\left(\frac{r^2}{2}\right)^{n/2}.$$ This implies that the area of the sphere of radius $r$ is $$V_n'(r)=2\times \frac{\pi^{n/2}}{\Gamma(n/2)}\times r^{n-1}.$$ Check it: $V'_3(r)=4\pi r^2.$