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I'm working on Exercise 25.5 from Munkres' Analysis on Manifolds, in his section about integration on manifolds. It asks to express the volume of $S^n(a)$ in terms of the volume of $B^{n-1}(a)$. Here $S^n(a)=\{x\in\mathbb{R}^{n+1}:\|x\|=a\}$, and $B^{n}(a)=\{x\in\mathbb{R}^{n}:\|x\|\leq a\}$. There is a hint to proceed as in an example in the book, where $v(S^2(a))$ is calculated by looking with the intersection of $S^2(a)$ with the plane $z=z_0$ with absolute value less than $a$, and then a polar parametrization is used. However, I don't think 'proceeding' as in this example will work, since then I have to use an $n$-dimensional spherical parametrization, which does not yield an expression in $B^{n-1}(a)$.

I understand that the recurrence should be $S^n(a)=2\pi aB^{n-1}(a)$, but I don't know how to prove this.

I'm looking for some help, since I don't understand how to proceed. I found a few posts online, but all had a geometrical approach; here I'm looking more for an analysis approach. Any help is much appreciated and if you need more context, let me know.

By the way, for the second part I have to show that $v(S^n(t))=Dv(B^{n+1}(t))$, which I want to solve as follows: we can look at $B^{n+1}(t)$ as $\cup_{r\in[0,1]}rS^n(t)$, where this union is disjoint, thus $$v(B^{n+1}(t)=\int_0^1v(rS^n(t))\mathrm{d}r=\frac{v(S^n(t))}{n+1};$$ taking the derivative then shows $v(S^n(t))=Dv(B^{n+1}(t))$. However, this does not feel as a good proof to me, since I don't know if the integral equality is justified. Also, I don't see how the first part helps here.

Descartes Before the Horse
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Václav Mordvinov
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  • You are using the very confusing notation $B^{n-1}$ in place of $B^n$ – Federico Dec 12 '18 at 18:50
  • You're right, it is a typo actually, I'll edit this directly. Thanks for noting – Václav Mordvinov Dec 12 '18 at 18:52
  • Do you know the measure theoretic result that there exists a unique measure $\sigma$ on $S^{n-1}$ such that Lebesgue measure $m$ on $\mathbb{R}^n$ can be expressed as a product measure $(\rho^{n-1} d\rho) \otimes \sigma$ where $\rho = \lVert x \rVert$? (And then this $\sigma$ agrees with the volume measure from the induced Riemannian metric on $S^{n-1}$. You can then also calculate $\sigma(S^{n-1})$ and $m(B^n)$ by applying this to $\int_{\mathbb{R}^n} \exp(-\lVert x \rVert^2) dm(x)$.) – Daniel Schepler Dec 12 '18 at 19:11
  • @DanielSchepler no, I've never heard of all these facts. I'm a second year undergraduate student currently doing a first course on analysis on manifold. – Václav Mordvinov Dec 12 '18 at 21:03
  • Related, possibly of interest. – Andrew D. Hwang Apr 13 '21 at 15:11

1 Answers1

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Parameterize $S^n(a)$ by $\varphi : (0, 2\pi) \times B^{n-1}(a) \to S^n(a)$ where \begin{equation} \varphi(\theta, x_1, \dots, x_{n-1}) = \left(r\cos\theta, r\sin\theta, x_1, \dots, x_{n-1}\right) \end{equation} where $r = \sqrt{a^2 - x_1^2 + \cdots + x_{n-1}^2}$. Then \begin{equation} D\varphi = \begin{bmatrix} -r\sin\theta & x_1/r\cos\theta & \cdots & x_{n-1}/r\cos\theta \\ r\cos\theta & x_1/r\sin\theta & \cdots & x_{n-1}/r\sin\theta \\ 0 \\ \vdots \\ 0 & & I_{n-1} \\ \vdots \\ 0 \end{bmatrix} \end{equation}

Let $(D\varphi)_i$ be the matrix obtained from $D\varphi$ by removing the $i$th row. Then $\det(D\varphi)_1 = r\cos\theta$, $\det(D\varphi)_2 = -r\sin\theta$, and $\det(D\varphi)_{k+2} = x_k$. The value of this last determinant can be obtained by adding $-\sin\theta / \cos\theta$ times the first row to the second and expanding by minors along the first column. Now the volume element is \begin{equation} V(D\varphi)^2 = \sum_{i=1}^{n+1} [\det(D\varphi)_i]^2 = r^2\sin^2\theta + r^2\cos^2\theta + x_1^2 + \cdots + x_{n-1}^2 = a^2 \end{equation} Hence by Fubini's theorem we have \begin{equation} v(S_n(a)) = \int_{[0, 2\pi] \times B^{n-1}(a)} V(D\varphi) = \int_0^{2\pi} \int_{B^{n-1}(a)} a = 2\pi a \cdot v(B^{n-1}(a)) \end{equation}

AbelianGrape
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