Nathalie Fields

Answered

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?

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?

Answer & Explanation

losnonamern

Expert

2022-07-17Added 12 answers

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 ${V}_{n}(k)$ and recognize that ${V}_{n}(k)={\int}_{0}^{k}{V}_{n-1}(x)dx$.

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.

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 ${V}_{n}(k)$ and recognize that ${V}_{n}(k)={\int}_{0}^{k}{V}_{n-1}(x)dx$.

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

2022-07-18Added 6 answers

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.

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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