# N-Dimensions

## 2-Dimensional Universes

See "The Planiverse" by A. K. Dewdney on my Book List.

## N-Dimensional Hyperspheres and Hyperballs

The volume of an n-dimensional hypercube is simply s^{n} where
s is the length of a side.

What is the volume of an n-dimensional hyperball?

First, let's define some familar geometric shapes.

- A circle is the 1-dimensional rim of 2-dimensional disk.
- A disk is the 2-dimensional area filling a circle.
- A sphere is the 2-dimensional surface of a 3-dimensional ball.
- A ball is the 3-dimensional volume filling a sphere.

In higher dimensions we use the terms hypersphere and hyperball, or, to be more specific, an n-sphere or n-ball.

- An n-sphere is a hypersphere with n dimensions.
- An n-ball is a hyperball with n dimensions.

We can redefine the familar shapes using our new terms.

- A circle is a 1-sphere.
- A disk is a 2-ball.
- A sphere is a 2-sphere.
- A ball is a 3-ball.

The n-content is the n-dimensional "area" or "volume" of a geometric shape. For example:

- The 1-content of a circle is its circumference.
- The 2-content of a disk is its area.
- The 2-content of a sphere (2-sphere) is its surface area.
- The 3-content of a ball (3-ball) is its volume.
- The 3-content of a 3-sphere (hypersphere) is its hyper-surface-area.
- The 4-content of a 4-ball (hyperball) is its hyper-volume.

Here is a table showing, for different dimensions, the n-content ("volume") of hyperballs and the boundary (n-1)-content ("surface area") of their corresponding hyperspheres:

Dimension (n) | Full Shape | Full n-Content ("volume") |
Boundary Shape | Boundary (n-1)-Content ("surface area") |
---|---|---|---|---|

2 | disk (2-ball) | π r^{2} |
circle (1-sphere) | 2π r |

3 | ball (3-ball) | (4/3)π r^{3} |
sphere (2-sphere) | 4π r^{2} |

4 | 4-ball | (1/2)π^{2}r^{4} |
3-sphere | 2π^{2}r^{3} |

5 | 5-ball | (8/15)π^{2}r^{5} |
4-sphere | (8/3)π^{2}r^{4} |

6 | 6-ball | (1/6)π^{3}r^{6} |
5-sphere | π^{3}r^{5} |

7 | 7-ball | (16/105)π^{3}r^{7} |
6-sphere | (16/15)π^{3}r^{6} |

Isn't it strange that the power of π increases by one only when the dimension increases by two?

In general, the n-content ("volume") of an n-dimensional hyperball is:

if n is even: | (1/(n/2)!)π^{n/2}r^{n} |

if n is odd: | (2^{n}((n-1)/2)!/n!)π^{(n-1)/2}r^{n} |

or (2^{(n+1)/2}/n!!)π^{(n-1)/2}r^{n} |

where n! = n(n-1)(n-2)... (factorial) and n!! = n(n-2)(n-4)... (double factorial).

In general, the boundary "surface area" ((n-1)-content) of an n-dimensional hyperball is the "volume" (n-content) multiplied by (n/r).

Using differential calculus, you can find the "surface area" ((n-1)-content) of an n-ball by differentiating its "volume" (n-content) with respect to the radius, r. This is a fun exercise for calculus beginners.

Using integral calculus, you can derive the formulas for n-dimensional balls based on the formulas for (n-1)-dimensional balls. The "volume" of an n-dimensional ball can be found by integrating the "surface area" of (n-1)-dimensional spherical shells from 0 to r, like the layers of an onion.