# 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 sn 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 r2 circle (1-sphere) 2π r
3 ball (3-ball) (4/3)π r3 sphere (2-sphere) 4π r2
4 4-ball (1/2)π2r4 3-sphere 2r3
5 5-ball (8/15)π2r5 4-sphere (8/3)π2r4
6 6-ball (1/6)π3r6 5-sphere π3r5
7 7-ball (16/105)π3r7 6-sphere (16/15)π3r6

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)!)πn/2rn if n is odd: (2n((n-1)/2)!/n!)π(n-1)/2rn or (2(n+1)/2/n!!)π(n-1)/2rn

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.