Volume of a n-dimensional Ball: Part 1

Volume of a n-dimensional Ball: Part 1

Comments 1 comment

1. Introduction

This post investigates an interesting question I had been thinking about: we all know that the area of a circle of radius r equals \pi r^2, and the volume of a sphere of radius r equals \frac{4}{3}\pi r^3. But what if we enter the territory of even higher dimensions? We should be able to say something about the “volume” of an n-dimensional sphere, where “volume” is interpreted as the equivalent measure of space in that many dimensions (e.g. area for 2D, regular volume for 3D). This topic will be broken into two posts due to its length, but will culminate in a complete proof I found of the known formula for the volume of n-dimensional spheres. Familiarity with Calculus I and II is recommended. Also, note that throughout this post, “sphere” will be replaced by “ball” to denote the fact that it is “filled in,” hence an n-dimensional ball is an n-dimensional manifold embedded in n-dimensional space. Without further ado, let us begin.

2. Preliminaries

To make things simple, the first thing we will do is mandate that all balls we consider will have a radius of 1. This is not a problem, because this unit ball can be scaled by any factor r to result in a ball of radius r, and the volume will simply be multiplied by r^n if there are n dimensions. From now on, denote the volume of an n-dimensional ball of radius 1 as V_n.

Before moving on to higher dimensions, we must first question how the 3-dimensional case is actually derived. That is, how do we know that V_3=\frac{4}{3}\pi? There are many ways to obtain this result, but we will use a calculus-based approach. Imagine a unit 3-D ball centered around the origin. We will apply the standard method of dividing this ball into infinitely many thin slices, each slice perpendicular to the x-axis, and then add up all these volumes via integration. Note that the slice located at any particular value of x is a 2-D circle with radius \sqrt{1 - x^2} by the Pythagorean Theorem, so its area is \pi (\sqrt{1 - x^2})^2. However, for our purposes, we will write this slightly differently, as V_2 (\sqrt{1 - x^2})^2 (remember, V_n is the volume of an n-D ball, so V_2 = \pi). The thickness of this slice is dx, so because the sphere’s radius is 1, we must integrate the area of the slices from x = -1 to x = 1. We have:

    \begin{equation*} $V_3=\int_{-1}^{1} V_2 (\sqrt{1 - x^2})^2 dx = V_2\int_{-1}^{1} (\sqrt{1 - x^2})^2 dx$ \end{equation*}

The integral can easily be evaluated as \frac{4}{3}, yielding the correct value of V_3 = \frac{4}{3}V_2 = \frac{4}{3}\pi. We now have a game plan: given that we know the value of V_n, we can recursively calculate the value of V_{n + 1} by setting up a similar integral.

3. Setup

Imagine an n-D unit ball in n-D space, centered on the origin. Because the space is n-dimensional, it has a total of n axes. Pick any one and call it the x-axis, this will be the only important one, because it is the one we will integrate over. Consider “slicing” our ball with a hyperplane perpendicular to the x-axis at some particular x value between -1 and 1. In our previous 3-D case, the slices were in the shape of 2-D circles: we can see that each slice in the general case will be shaped like (n - 1)-D balls, with an infinitesimal thickness dx along the x-axis. So the volume of each slice equals dx times the “lower dimensional” volume of the (n - 1)-D ball. But what is the radius of each smaller ball? Consider an arbitrary point on the edge of one of the (n - 1)-D balls, and draw a radius from the center of the ball to that point on the edge. Call this radius R. Because the (n - 1)-D ball is located on a slice at some particular x_0 on the x-axis, any length within it (including the drawn radius) must be perpendicular to the x axis itself. So by the Pythagorean Theorem, the distance from the origin (center of the n-D ball, not the (n - 1)-D ball) to the selected point on the boundary of the (n - 1)-D ball equals \sqrt{x_{0}^2+R^2}. However, because that point is also located on the boundary of the entire n-D ball itself, this distance is simply equal to the radius of the n-D ball, 1. Hence, we find R=\sqrt{1 - x_{0}^2}. Now that we know the radius of the slice, we can calculate its volume: if the radius was 1, the volume would be V_{n - 1}, but we must scale the radius by a factor of R, so its volume is:

    \begin{equation*} $V_{n - 1} R^{n - 1} = V_{n - 1} \left(\sqrt{1 - x_{0}^2}\right)^{n - 1}$ \end{equation*}

Going back to the larger problem, we see that all we have to do is integrate this smaller volume * dx over the x-axis from -1 to 1:

    \begin{equation*} $V_n = \int_{-1}^{1} V_{n - 1} \left(\sqrt{1 - x^2}\right)^{n - 1} dx = V_{n - 1}\int_{-1}^{1} \left(\sqrt{1 - x^2}\right)^{n - 1} dx$ \end{equation*}

We note something very interesting, to get from V_{n - 1} to V_n, we must simply multiply by a specific constant term dependent on n. From now on, denote this term as

    \begin{equation*} $a_n = \int_{-1}^{1} \left(\sqrt{1 - x^2}\right)^{n - 1} dx$ \end{equation*}

so we have V_{n} = a_{n}V_{n - 1}. We will conclude the first part of this post with the following observation: we may substitute x = \sin(t) and dx = \cos(t)dt in the above integral for a_n; when we do so, we note that

    \begin{equation*} $\left(\sqrt{1 - x^2}\right)^{n - 1} = \left(\sqrt{1 - \sin^2(t)}\right)^{n - 1}=\cos(t)^{n - 1}$ \end{equation*}

Meanwhile, the bounds (-1, 1) turn into (-\frac{\pi}{2}, \frac{\pi}{2}). Thus, we can write

    \begin{equation*} $a_n = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{n}(t) dt$ \end{equation*}

Feel free to use the recursion to explore and calculate a few higher values of V_n. In the second installment, I will investigate deeper properties of this integral in order to prove the known explicit formula for V_n.

One thought on “Volume of a n-dimensional Ball: Part 1”

Leave a Reply

Your email address will not be published. Required fields are marked *