Let me first state the statement which I want to prove (encountered while studying "Geometry of Number"):

Suppose $A$ is a convex, measurable, compact and centrally symmetric subset of $\mathbb{R}^n$ where $\mathbf{x}=(x_1, \ldots, x_r, x_{r+1}, \ldots, x_n)\in A$ iff $|x_i|\leq 1$ for $1\leq i \leq r$ and $x_{r+1}^2 + x_{r+2}^2 \leq 1, \ldots, x_{n-1}^2 + x_{n}^2 \leq 1$

Then $\text{vol}(A) = 2^r \pi^{\frac{n-r}{2}}$

The definition of various terms used above :

  • $A\subset \mathbb{R}^n$ is called a convex set if $\mathbf{x}$ and $\mathbf{y}$ are in $A$ then so is the entire line segment joining them.

  • Measurable refers to Lebesgue measure in $\mathbb{R}^n$; the Lebesgue measure $\text{vol}(A)$ coincides with any reasonable intuitive concept of n-dimensional volume, and Lebesgue measure is countably additive.

  • Centrally symmetric means symmetric around $\mathbf{0}$: if $\mathbf{x} \in A$ then so is $-\mathbf{x}$.

I have no idea about how to approach this problem.

Edit: It is also given that $n-r$ is an even number.

  • 2
    $\begingroup$ Well, the first $r$ factors are just numbers between $-1$ and $1$, so your set is a product of an $r$-cube of side 2 with "the stuff with indices greater than $r$". Since the volume of a product is the product of the volumes, this accounts for the factor of $2^r$, and all you have to think about is the last $n-r$ terms. That's still something of a mess, but it's not too awful, and you can do index-shifting to make the whole description a little simpler. $\endgroup$ Jul 12, 2016 at 12:57
  • $\begingroup$ Aren't the last $n-r$ or ${n-r\over 2}$ conditions just disks of radius $1$? $\endgroup$ Jul 12, 2016 at 13:45
  • $\begingroup$ It should say $x_{r+j}^2+x_{r+j+1}^2\leq 1$ when $j $ is ODD. The set $A$ is then $[-1,1]^r\times D^{(n-r)/2}$ where $D=\{(x,y): x^2+y^2\leq 1\}.$ $\endgroup$ Jul 12, 2016 at 15:44
  • $\begingroup$ @user254665 this is what has been done by Yaddle below. $\endgroup$ Jul 12, 2016 at 15:57

1 Answer 1


To keep things simple we will do some induction: Set $A_r = \{(x_1, \dots, x_r) \in \mathbb R^r : \forall i \in \{1,\dots, n\}: \vert x_i \vert \leq 1\}$. Consider the integral $\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r)$, we will prove by induction that $$\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) = 2^r.$$ For $n = 1$ we get $$\int_{A_1} 1 \ d x_1 = \int_{-1}^1 1\ dx_1 = 2.$$ Now let the statement hold for $r \in \mathbb N$. Then we get for $r + 1$ with Fubini that $$\int_{A_{r+1}} 1 \ d\lambda(x_1, \dots, x_{r+1}) = \int_{-1}^1 \bigg(\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) \bigg)dx_{r+1} = \int_{-1}^1 2^r\ dx_{r+1} = 2^{r +1}.$$ Now to the second part. Let $k := n - r$ and $B_{k} = \{(x_{r+1}, \dots, x_n) \in \mathbb R^{n-r} : x_{r+1}^2 + x_{r+2}^2 \leq 1, \ldots, x_{n-1}^2 + x_{n}^2 \leq 1 \}$ for even $k$. Now we prove by induction that $$\int_{B_k} 1 \ d\lambda(x_{r+1}, \dots, x_n) = \pi^{k/2}.$$ For $k = 2$ we get by using polar coordinates $$ \int_{B_2} 1\ d\lambda(x_{r+1}, x_{r+2}) = \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \int_0^1 2\pi r\ dr = \pi.$$ Now let the statement hold for $k \in \mathbb N$ even. Then we get for $k + 2$ with Fubini that \begin{align*}\int_{B_{k+2}} 1\ d\lambda(x_{r+1}, \dots, x_{n + 2}) &= \int_{B_2} \bigg( \int_{B_{k}} 1\ d\lambda(x_{r+1}, \dots, x_n)\bigg) d\lambda(x_{n+1}, x_{n + 2}) = \int_{B_2} \pi^{k/2}\ d\lambda(x_{n+1}, x_{n + 2}) \\ &= \pi^{k/2} \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \pi^{k/2} \pi = \pi^{(k + 1)/2}\end{align*} All together we can get know with Fubini \begin{align*}\operatorname{vol}(A) &= \int_A 1\ d\lambda(x_1, \dots, x_n) = \int_{A_r} \bigg(\int_{B_k} 1\ d\lambda(x_{r+1}, \dots, x_n) \bigg) d\lambda(x_1, \dots, x_r) \\ &= \int_{A_r} \pi^{k/2}\ d\lambda(x_1, \dots, x_r) = \pi^{k/2} \int_{A_r}1 \ d\lambda(x_1, \dots, x_r) = \pi^{k/2} 2^r.\end{align*} And that was the result we were aiming for.

The "physics" version: Denote the unit disk by $D = \{x \in R^2 : x_1^2 + x_2^2 \leq 1\}$. Then $A = [-1,1]^r \times D^{(n - r)/2}$. Show like above that $\operatorname{vol}([-1,1]) = 2$ and $\operatorname{vol}(D) = \pi$. Then we got $$\operatorname{vol}(A) = \operatorname{vol}([-1,1]^r \times D^{(n - r)/2}) = \operatorname{vol}([-1,1])^r \operatorname{vol}(D)^{(n - r)/2} = 2^r \pi^{(n - r)/2}.$$ This prove makes use of the fact you can write $A$ as $A = [-1,1]^r \times D^{(n - r)/2}$ and the relation between the one-dimensional Lebesgue measure and the multi-dimensional Lebesgue measure.

Hope it helps :)

  • $\begingroup$ Using your method I can directly prove the general formula that when $|x_1| + \ldots + |x_r| + 2(\sqrt{x_{r+1}^2 + x_{r+2}^2}+\ldots + \sqrt{x_{n-1}^2 + x_{n}^2}) \leq n$ then $\text{vol}(A) = \frac{n^n}{n!} 2^r (\frac{\pi}{2})^{k/2}$ $\endgroup$ Jul 12, 2016 at 13:52
  • $\begingroup$ Yes, you can modify the result by using the same arguments. $\endgroup$
    – Yaddle
    Jul 12, 2016 at 13:56
  • $\begingroup$ There should exist a simple argument for the case stated in question, since that case is used as motivation to think about general case, which is proved later on. $\endgroup$ Jul 12, 2016 at 13:56
  • $\begingroup$ The essence is that you can split the integral over $A$ into integrals that don't depend on one another, since you can cut $A$ into simple one dimensional pieces. And that you can calculate the integral over a cuboid by multiplying side lengths. I just wrote in down with mathematical precision. A physician for example would write down the same result much shorter. $\endgroup$
    – Yaddle
    Jul 12, 2016 at 14:01
  • $\begingroup$ Can you please also add how a physicist would write down the result? Can you extend the hint provided above by @JohnHughes? $\endgroup$ Jul 12, 2016 at 14:03

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