Nathalie Fields

2022-07-16

Volume of Generalized Tetrahedron in ${R}^{n}$.
I'm having difficulty finding the volume of a tetrahedron in ${\mathbb{R}}^{n}$.
Find the volume of a generalized tetrahedron in ${\mathbb{R}}^{n}$ bounded by the coordinate hyperplanes and the hyperplane ${x}_{1}+{x}_{2}+...+{x}_{n}=1$
In two dimensions, we have ${\int }_{0}^{1}1-{x}_{1}d{x}_{1}$. In three dimensions, I got something like ${\int }_{0}^{1}{\int }_{0}^{1-{x}_{1}}1-{x}_{2}d{x}_{2}d{x}_{1}$.
I am off to a good start?

losnonamern

Expert

Step 1
Your approach is fine, but the allowable range in ${x}_{3}$ is $1-{x}_{1}-{x}_{2}$, so that should be your integrand. It is probably easier to define the n-volume of a k sided simplex as ${V}_{n}\left(k\right)$ and recognize that ${V}_{n}\left(k\right)={\int }_{0}^{k}{V}_{n-1}\left(x\right)dx$.
Step 2
Now each integral is a single one. If you do the first few, you will see a pattern emerge, which you can prove by induction.

Paxton Hoffman

Expert

Step 1
It is more accurate to write for two dimensions like this:
${\int }_{0}^{1}d{x}_{1}{\int }_{0}^{{x}_{1}}d{x}_{2}$
Step 2
For third dimension it will be:
${\int }_{0}^{1}d{x}_{1}{\int }_{0}^{{x}_{1}}d{x}_{2}{\int }_{0}^{{x}_{2}}d{x}_{3}$
So you can easily write the expression for higher dimensions. My advise is to write the domain bound by hyperplanes more carefully. Your way is too difficult for further integration I guess.

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