Find the volume of the set $S=\{x=(x_1,x_2,\cdots,x_n)\in \Bbb{R}^n:0\le x_1\le x_2\le \cdots \le x_n \le 1\}$

Question

Let $$S=\{x=(x_1,x_2,\cdots,x_n)\in \Bbb{R}^n:0\le x_1\le x_2\le \cdots \le x_n \le 1\}$$

Find the volume of the set $S$.

I tried writing it as a multiple integral but it got complicated.

Answer 1

You certainly can do this with a multiple integral; but here's another approach.

Imagine picking $n$ values, $X_1, \dots, X_n$ randomly, uniformly, and independently from $[0,1]$.

Then the probability that $X_1\le X_2 \le \cdots \le X_n$ is $\frac{1}{n!}$ (since there are $n!$ orderings of these variables).

But also this probability should equal the proportion of the volume of your region relative to the $n$-dimensional unit hypercube.

Answer 2

We can define a sequence of functions as follows:

$$V_1(x)=\int_0^x1dt=x\\ V_{n+1}(x)=\int_0^xV_n(t)dt$$

Then $V_n(1)$ will be the volume in $\Bbb R^n$. It may look a bit intimidating but really it is just

$$V_n(1)=\int_0^1\int_0^{x_n}\cdots\int_0^{x_3}\int_0^{x_2}dx_1dx_2\cdots dx_{n-1}dx_n$$

Let's try to show that $V_n(x)=\frac{x^n}{n!}$ for all $n$, through induction. The case $n=1$ is true. Assuming there exists a $k$ for which it is true,

$$V_{k+1}(x)=\int_0^xV_k(t)dt=\int_0^x\frac{t^n}{n!}dt=\frac{x^{n+1}}{(n+1)!}$$

So it follows that it is true for $k+1$ as well. Then the volume in $\Bbb R^n$ is

$$V_n(1)=\frac1{n!}$$

as paw88789's answer also stated.

Answer 3

You are asking to compute the volume of an $n$-dimensional pyramid. It is well known that the volume of a pyramid $P$ (in general of any cone) is: $$ V(P) = \frac{\text{volume of the base} \times \text{height}}{n}. $$ Since the base is again a $(n-1)$-dimensional pyramid, and the height is always one, you obtain the result by induction: $\frac{1}{n} \frac{1}{n-1}\dots = \frac{1}{n!}$.

The above formula can be easily proven by Cavalieri principles (or Fubini theorem). The measure of a set can be computed as the integral of the measure of slices. If you slice a cone with a plane parallel to the base you get a figure which is congruent to a rescalement of the base. More precisely, if a cone $C$ has base $B$ and height $h$ you have: $$ V(C) = \int_0^h \left(\frac{t}{h}\right)^{n-1}V(B)\, dt = \frac{h}{n} V(B). $$

Answer 4

For $n=1$, $S_1$ is a unit line segment and its volume is $V_1 = 1$.
For $n=2$, $S_2$ is a rectangular triangle with $S_1$ as a base and a unit height and $V_2 = \tfrac 12V_1 = \tfrac 12$.
For $n=3$, $S_3$ is a pyramid with $S_2$ in a base and a unit height and $V_3 = \tfrac 13 V_2 = \tfrac 16$.
....
For any $n>1$, $S_n$ is a simplex with $S_{n-1}$ as a base and a unit height and $V_n = \tfrac 1n V_{n-1} = \tfrac 1{n!}$.