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 Area-Volume Formulas for N-Dimensional Spheres and Balls

In two dimensions there are the formulas that the area of disk enclosed withing a circle of radius R is πR2 and the circumference of that circle is 2πR. In three dimensions the formula for the volume of the ball enclosed within a sphere of radius R is (4/3)πR3 and for the area of the sphere is 4πR2. Note that the terminology is that a circle of radius R or a sphere of radius R apply to geometric figures whose points lie a distance of exactly R from their centers. The figures whose points lie a distance which is less than or equal to R are called a disk for two dimensions and a ball for three dimensions. In the material below a disk is considered a two dimensional ball. The purpose of this material is to derived the formulas for the volume n-dimensional balls and then use those to derive the formula for the area of the (n-1)-dimensional sphere which surrounds an n-dimensional ball.

The terminology can be extended to one dimension. A one dimensional ball of radius R is a line segment of length 2R and its zero dimensional sphere consists of two points separated by a distance of 2R.

The area of a disk can be found as the limit of a sequence of approximations in which the disk is covered by a set of rectangles as shown in the diagrams below.

In the above construction the vertical axis of the disk is divided into m equal intervals. The width of a rectangle is the width of the disk at that height. As the subdivision of the vertical axis of the disk becomes finer and finer the sum of the areas of the rectangles approaches a limit which is called the area of the disk.

The process can be represented algebraically by noting that the equation of the circle which encloses the disk can be expressed in terms of the horizontal and vertical distances from the center, x and y, as x2 + y2 = R2. Thus at a height y the width of the disk is 2(R2-y2)1/2. The limit can be expressed as an integral; i.e.,

#### V2(R) = ∫-RR2(R2-y2)1/2dy

The evaluation of the integral can be greatly simplified by noting that the cross width of the disk can be expressed as 2Rcos(θ) where θ is the angle shown in the figure below. Note also that y=Rsin(θ) so dy=Rcos(θ)dθ.

Thus:

#### V2(R) = ∫-π/2π/22Rcos(θ)Rcos(θ)dθ = ∫-π/2π/22R2cos2(θ)dθ or, by symmetry, V2(R) = 2∫0π/22R2cos2(θ)dθ

The final result is then

#### V2(R) = 4R2∫0π/2cos2(θ)dθ V2(R) = 4R2[(1/2)(θ + sin(θ)cos(θ))]0π/2 V2(R) = 4R2(π/4) = πR2

The result is trivial but the process can be generalized to n dimensions. First, one of the variables of the equation for an n-dimensional ball is selected, say z. At each value of z there is a cross-section which is an (n-1)-dimensional ball. In the case considered above the cross-section of the disk was a line segment of length 2r (a one dimensional ball) where r is the cross-section radius which can be expressed as Rcos(θ). Integration over θ from -π/2 to π gives the n-dimensional volume. This means that the volume formulas can be found iteratively; i.e.,

#### Vn(R) = 2∫0π/2Vn-1(Rcos(θ))Rcos(θ)dθ

Thus, knowing that V2(R)=πR2

#### V3(R) = 2∫0π/2V2(Rcos(θ))Rcos(θ)dθ = 2∫0π/2πR2Rcos(θ)dθ =2∫0π/2πR3cos3(θ)dθ = 2πR3∫0π/2cos3(θ)dθ

Since ∫cos3(θ)dθ= sin(θ) - (1/3)sin3(θ)
0π/2cos3(θ)dθ = 1 - 1/3 = 2/3
this means:

#### V3(R) = 2πR3(2/3) = (4/3)πR3

For n=4 the iteration scheme gives:

#### V4(R) = 2∫0π/2V3(Rcos(θ))Rcos(θ)dθ = 2∫0π/2(4/3)πR3cos3(θ)Rcos(θ)dθ = (8/3)πR4∫0π/2cos4(θ)dθ

The indefinite integral ∫cos4(θ)dθ is equal to (3/8)θ + (3/8)sin(θ)cos(θ) + (/4)sin(θ)cos3(θ). On evaluation of this expression between 0 and π/2 all terms involving the product of sin(θ) and cos(θ) reduce to 0. Therefore

#### V4(R) = (8/3)πR4[(3/8)π/2] = (1/2)π2R4

Likewise V5(R) reduces to:

Furthermore:

and

## The Area of the (n-1)-Dimensional Sphere Bounding an n-Dimensional Ball

The area of the boundary of an n-dimensional ball is found simply by differentiating the volume formula. Thus for the circumference of the circle bounding a a disk of radius R one obtains d(πR2)/dR = 2πR. For a three dimensional ball of radius R the area of the bounding sphere is d((4/3)πR3)/dR = 4πR2. When this procedure is applied to the one dimensional ball, a line segment of length 2R, the result is that the area of the bounding end points is d(2R)/dR = 2, the number of points. This gives an interesting insight; that the zero dimensional analogue of volume is the number of points, the cardinality of the boundary set.

For a 4 dimensional ball the result is that the boundary 3D ball has an area (volume in this case) of d((1/2)π2R4)/dR = 2π2R3. Note that because of the second power of π this does not correspond to a simple number of 3D balls.

For n=5 the area of the 4D boundary is d((8/15)π2R5)/dR = (8/3)π2R4. In this case the boundary is equivalent to 16/3 4D balls.

## Summary of the Volume-Area Relationships by Dimension

DimensionVolumeBounding
Area
12R2
2πR22πR
3(4/3)πR34πR2
4(1/2)π2R42R3
5(8/15)π2R5(8/3)π2R4
6(1/6)π3R6π3R5
7(16/105)π3R7(16/15)π3R6

## Generalization

The pattern is of the form

#### Vn = Cnπ[n/2]Rn

where [n/2] is the integer value of (n/2). The area formula then takes the form

#### An = nCnπ[n/2]Rn-1

The previously defined iteration formula

The integral

#### If (n+1)/2 is equal to an integer m then Γ((n+1)/2)=(m-1)!. If (n+1)/2=m+½ then Γ((n+1)/2)=π½(2m-1)!!/2m

where the notation p!! denotes p(p-2)(p-4)… down to 2 if p is even and down to 1 if p is odd.

Thus the value of Cn can be determined iteratively, though messily, for any value of n.

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