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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 $n$?

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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 at 1:19

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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, $$V(T_n(1))=\frac1{n!}.$$

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thank you very much !! –  Math can be Fun May 1 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 at 9:13
    
ِDid you calculate Jacobian of this change of variable? –  user91500 May 1 at 9:23
    
I did it for the case $n=3$ which is $1$ –  Math can be Fun May 1 at 9:24
    
and $n=2$ also . –  Math can be Fun May 1 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.

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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 $(1-x)^{n-1}$.

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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 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 at 1:14
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I have fixed my response, which is now a more fleshed out version of Ross'. –  Ben Whitney Apr 26 at 1:26

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