Volume calculation in R^n I am interested in the following calculation, it seemed simple enough but I was having some trouble. I am not a mathematician so I wouldn't know if there already exists a set of methods for dealing with this kind of problem. Suppose you have a region of space in $\mathbb{R}^n$ defined by the set:
$$\{\textbf{x} \in \mathbb{R}^n : \textbf{a} \cdot \textbf{x} ≤ b\; \mathrm{and} \; \textbf{x} ≥ 0 \}$$
where $\textbf{a} ≥ 0$ and $b > 0$. In other words, it's the region bounded between the hyperplane $\{\textbf{x} \in \mathbb{R}^n : \textbf{a} \cdot \textbf{x} = b \}$ and the coordinate axes. I am interested in calculating the volume of this region. This is simple enough to do for $n = 2$ or $n = 3$ but I have not been able to calculate a general formula for arbitrary $n$. Anybody know how to do this? Thanks in advance.

The hyperplane $\textbf{a} \cdot \textbf{x} = b$ intersects the $k$-th axis at $b/a_k$. Suppose you computed the $n-1$ dimensional volume $V_{n-1}$, by integrating over the first $n-1$ coordinates with $x_n=0$. The intersections between your region and a generic hyperplane $x_n=c$, perpendicular to the $n$-th axis, are $n-1$ dimensional regions, similar between them and whose linear dimensions decrease linearly going from $x_n=0$ (when the volume of the intersection is $V_{n-1}$) to $x_n=b/a_n$ (when the volume of the intersection is $0$). The intersection volume at a generic $x_n$ is then $f(x_n)=V_{n-1}(1-x_n a_n/b)^{n-1}$ and you can find $V_n$ by integrating this from $0$ to $b/a_n$:
$$
V_n=\int_0^{b/a_n}\!\!\!\!f(x_n)\,dx_n
=V_{n-1}\int_0^{b/a_n}\!\!\!\!(1-x_n a_n/b)^{n-1}dx_n
=V_{n-1}{b\over a_n}\int_0^1t^{n-1}dt
=V_{n-1}{b\over a_n}{1\over n}.
$$
Starting with $V_1=b/a_1$ one then gets:
$$
V_n={1\over n!}{b^n\over a_1\dots a_n}.
$$
