Volume of a region?

Question

It is my intuition that the volume of the solid such that $0\leq x_1 \leq x_2 \ldots \leq x_n \leq T$ is $\frac{T^n}{n!}$. Can someone confirm/deny and/or supply proof? Thanks!

Answer 1

For $T > 0$, let $V_n(T)$ denote the volume of the region $$\Sigma_n(T) := \{(x_1,\ldots,x_n)\in \Bbb R^n : 0 \le x_1 \le \cdots \le x_n \le T\}.$$

By taking cross-sections of $\Sigma_n(T)$ perpendicular to the $x_n$-axis, we find

$$V_n(T) = \int_0^T \int_{\Sigma_{n-1}(t)} dx^{n-1}\, dt = \int_0^T V_{n-1}(t)\, dt, \quad n > 1.$$

Note $V_1(T) = \operatorname{Vol}([0,T]) = T$. If $V_n(T) = \frac{T^n}{n!}$ for some $n > 1$, then

$$V_{n+1}(T) = \int_0^T V_n(t)\, dt = \int_0^T \frac{t^n}{n!}\, dt = \frac{1}{n!}\cdot \frac{T^{n+1}}{n+1} = \frac{T^{n+1}}{(n+1)!}.$$

By induction, $V_n(T) = \frac{T^n}{n!}$ for all $n\ge 1$.

Answer 2

The volume of the solid is $$\int_0^T\int_0^{x_n}\int_0^{x_{n-1}}\cdots\int_0^{x_2}1dx_1\cdots dx_{n-2}dx_{n-1}dx_{n}$$ This easily succumbs to induction.