Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $T_n = \{ x_i \geq 0 : x_1 + ... + x_n \leq 1 \} $.I know $T_n$ is tetrahedron. MY question: How can I compute the volutme of $T_n$ for every $n$?

share|cite|improve this question
Because my answer will be deleted soon I will answer your question about triple integrals here. For $T_{3}$ the integral would be $\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-(y+z)}1 \, \mathrm{d} x \, \mathrm{d} y \, \mathrm{d} z$. In $n$ dimensions you would require a repeated integral over $n$ variables. – Ben Whitney Apr 26 '14 at 1:19
up vote 7 down vote accepted

First suppose that we want to find the volume of $T_n(a)$, then by change of variables $X=aU$, you can see that we have $V(T_n(a))=\color{red}{a^n}V(T_n(1))$.

Since $x_1+\cdots+x_n\le1$ if and only if $x_n\le1$ and $x_1+\cdots+x_{n-1}\le 1-x_n$, we have $$\begin{align}V(T_n(1))&=\int_{x_n\le 1}\left(\int_{x_1+\cdots+x_{n-1}\le1-x_n}dx_1\cdots dx_{n-1}\right)dx_n\\ &=V(T_{n-1}(1))\int_{x_n\le1}\color{red}{(1-x_n)^{n-1}}dx_n=\frac1 nV(T_{n-1}(1))\end{align}$$ The numbers $V(T_n(1))$ satisfy in the above recursion formula, so $$V(T_n(1))=\frac1{n!}.$$

One another way is to consider the following integral

$$I=\int_{T_n(a)}e^{-(x_1+\cdots+x_n)}dx_1\cdots dx_n$$ Since $V(T_n(a))=a^nV(T_n(1))$, $$I=\int_0^\infty e^{-a}dV(T_n(a))=V(T_n(1))\int_0^\infty n a^{n-1}e^{-a}da=n!V(T_n(1))$$ But also we have $$I=\left(\int_0^\infty e^{-x}dx\right)^n=1$$ Hence, $$V(T_n(1))=\frac1{n!}.$$

share|cite|improve this answer
thank you very much !! – Math can be Fun May 1 '14 at 8:32
I got the answer to the other question :). plus the other question is easy. This one is harder. I have one question: earlier today I was thinking this problem can be solved maybe by using a change of variables? What do you think about using $y_i = \sum_{j=1}^i x_j $ ? – Math can be Fun May 1 '14 at 9:13
ِDid you calculate Jacobian of this change of variable? – MATHEMATIKER May 1 '14 at 9:23
I did it for the case $n=3$ which is $1$ – Math can be Fun May 1 '14 at 9:24
and $n=2$ also . – Math can be Fun May 1 '14 at 9:30

Hint: The general rule is that the $n$-volume of a simplex is $\frac 1n$ times the $ (n-1)$ volume of the base times the height, which you can prove by integration. The height is $1$, so you have a recurrence.

share|cite|improve this answer

Try using induction. $T_{1}$ is just in the interval $[0,1]$, which has volume (length) $1$. $T_{2}$ is a right triangle with volume (area) $1/2$. Now imagine the case for $n=3$. When we slice the tetrahedron $T_{3}$ at some height $z\in[0,1]$, we get a cross section that looks like $T_{2}$. But as $z$ gets bigger, the cross section gets smaller. Try to convince yourself that in general we have $$ v(T_{n})=\int_{0}^{1}(1-x)^{n-1}v(T_{n-1})\,\mathrm{d}x=\frac{1}{n}v(T_{n-1}) $$ Then, using the fact that $v(T_{1})=1$, we have $v(T_{n})=1/n!$. Note that we scale by $(1-x)^{n-1}$ instead of by $1-x$ because it is the linear dimensions of the $T_{n-1}$ slice that scale by $1-x$, which translates to the $\mathbb{R}^{n-1}$ volume of the slice scaling by $(1-x)^{n-1}$.

share|cite|improve this answer
Thanks Ben! I have one question. Is there a way to solve this problem using triple integrals? and maybe a change of variables? – Math can be Fun Apr 26 '14 at 1:03
Unfortunately I answered too hastily and this approach is wrong! I apologize. I believe Ross' answer is correct. For $T_{3}$ the integral would be $\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-(y+z)}1 \, \mathrm{d} x \, \mathrm{d} y \, \mathrm{d} z$. – Ben Whitney Apr 26 '14 at 1:14
I have fixed my response, which is now a more fleshed out version of Ross'. – Ben Whitney Apr 26 '14 at 1:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.