N-Dimensions

2-Dimensional Universes

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

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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 terms. A circle is the 1-dimensional rim (boundary) of 2-dimensional disk. A sphere is the 2-dimensional surface (boundary) of a 3-dimensional ball. Hyperspheres (n-spheres) and hyperballs (n-balls) can have different numbers of dimensions. A regular sphere is a 2-sphere. A regular ball is a 3-ball. A circle can also be called a 1-sphere. A disk can also be called a 2-ball.
The n-content is the n-dimensional "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 (a 2-sphere) is its surface area,
the 3-content of a ball (a 3-ball) is its volume,
the 3-content of a 3-sphere is its hyper-surface-area, and
the 4-content of a 4-ball 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	2π2r3
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.