# Calculating the Volume of a Triangular Pyramid in a Hard Way

| Comments |

## Background

There is an easy way of calculating the volume of $\{(x,y,z) \in \R^3 \mid 0 \le x,y,z \le t, x + y + z \le t\}$: just consider the permutation of $x,y,z$.1 This can be easily generalized to $n$ dimension.

## Another way using multiple integrals

$\begin{split} & \text{Let } A:= \left\{ (x_1,\ldots,x_n) \in \R^n \,\middle|\, 0 \le x_i \le 1 \,\forall 1 \le i \le n, \sum\limits_{i = 1}^n x_i \le 1 \right\}. \\ & \text{Volume of } A \\ =& \idotsint\limits_A \ud x_n \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 1} x_i} \ud x_n \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 2} x_i} \left(t - \sum\limits_{i = 1}^{n - 1} x_i\right) \ud x_{n - 1} \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 2} x_i} x_{n - 1} \ud x_{n - 1} \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 3} x_i} \frac{1}{2!} \left(t - \sum\limits_{i = 1}^{n - 2} x_i\right)^2 \ud x_{n - 2} \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 3} x_i} \frac{x_{n - 2}^2}{2!} \ud x_{n - 2} \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \int_{0}^{t} \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 4} x_i} \frac{1}{3!} \left(t - \sum\limits_{i = 1}^{n - 3} x_i\right)^3 \ud x_{n - 3} \cdots \ud x_3 \ud x_2 \ud x_1 \\ & \vdots \\ =& \int_{0}^{t} \frac{(t - x_1)^{n - 1}}{(n - 1)!} \ud x_1 \\ =& \frac{t^n}{n!} \end{split}$

## Why use this method?

To calculate its centre of mass.

$\begin{split} & \text{First component of its centre of mass} \\ =& \frac{n!}{t^n} \idotsint\limits_A x_1 \ud x_n \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \frac{n!}{t^n} \int_{0}^{t} x_1 \int_{0}^{t - x_1} \int_{0}^{t - x_1 - x_2} \cdots \int_{0}^{t - \sum\limits_{i = 1}^{n - 1} x_i} \ud x_n \cdots \ud x_3 \ud x_2 \ud x_1 \\ =& \frac{n!}{t^n} \int_{0}^{t} x_1 \,\frac{(t - x_1)^{n - 1}}{(n - 1)!} \ud x_1 \\ =& \frac{n!}{t^n} \left( \left. -x_1 \,\frac{(t - x_1)^{n - 1}}{(n - 1)!} \right|_{0}^t + \int_{0}^{t} \frac{(t - x_1)^n}{n!} \ud x_1 \right) \\ =& \frac{n!}{t^n} \left( 0 + \frac{t^{n + 1}}{(n + 1)!} \right) \\ =& \frac{t}{n + 1} \end{split}$

By symmetry, we conclude that the center of mass is $\left(\frac{t}{n + 1}, \frac{t}{n + 1}, \ldots, \frac{t}{n + 1}\right) \in \R^n$.

1. Simplex. (2015, March 29). In Wikipedia, The Free Encyclopedia. Retrieved 15:34, April 6, 2015, from http://en.wikipedia.org/w/index.php?title=Simplex&oldid=654074423