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Topic: Volumes in nD Using Basic High School Geometry
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Evelyn Sander

Posts: 187
Registered: 12/3/04
Volumes in nD Using Basic High School Geometry
Posted: Aug 25, 1993 4:17 PM
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This is a description of a geometric means to calculate the volume of
a n-ball inscribed in a n dimensional hypercube. In two dimensions we
know that the disk inscribed in the unit square has radius 1/2 and
therefore area pi*(1/2)^2=pi/4. In three dimensions, the ball
inscribed in the unit cube again has radius 1/2, where sqrt means take
the square root. Thus the volume of this ball is 4/3*pi*(1/2)^3=pi/6.
Paul Burchard, a postdoc at the Geometry Center, showed me how to
extend these results to n dimensions (denoted R^n) without using more
than basic high school geometry and a few pictures. The extension
turns out to be a recursive relation based on the two and three
dimensional results: the n-ball inscribed in the unit hypercube has
volume equal to pi/(2*n) times the volume of the (n-2)-ball. Perhaps
this gives a good way to introduce a high school geometry course to
higher dimensional spaces. In the course of reading the article,
please see the associated figures, available from the Forum by anonymous
ftp in the directory hs.geometry.article

By describing the ball as a cone over an object and using
some general properties of volumes of cones, I will be
able to find the volume of the ball. Here is a relationship
between spheres and solid balls using cones: By
definition, a cone over an object (not intersecting the
origin) consists of all the lines connecting that object
to the origin. A closed n-ball is the boundary and
interior of an (n-1)-sphere. More precisely, the sphere
S(n-1)={x in R^n:|x|=1}, and the ball B(n)={x in
R^n:|x|<=1}={t*x:x in S(n-1), t in [0,1]}. Thus the
n-ball is actually a cone over the (n-1)-sphere. Thus to
find the volume of the ball (Vol(n)), we need an equation
for the volume of a cone over a sphere. Note that in fact
Vol(n) is the volume of the ball of radius one, when really
we wanted the volume of a ball of radius 1/2. This means we
need to divide by 2^n at the end of the calculation.

We can think of the cone over the (n-1)-sphere as a union of
cones over infinitesimal n-1 dimensional cubes. Note
that the volume of a cone over a flat object only depends on
the distance from the plane of the object to the origin
(see Figure 1). The volume varies linearly with the
distance from the object to the origin, since only
varying one dimension makes the volume change linearly.
As an analogy, think of the cube (see Figure 2).

The n dimensional cube consists of exactly n identical cones over
n-1 dimensional cubes. To see this, note that a vertex V of
a cube intersects exactly n faces and has n faces disjoint
from it. Almost every point of the cube is in exactly one
cone between a disjoint face and V, the cube consists of
the union of cones from each of the n disjoint faces to V
(see Figure 3).

Given an arbitrary point P in the cube, draw a ray from V
through P. A point on this line intersects a disjoint
face. This face is unique for all but a set of zero volume.
Here is a slightly technical proof: Assume that we are
discussing the unit cube. If we make V the origin, and all
the other vertices vectors consisting of 0's and 1's,
then the cube is contained entirely in the first
quadrant. The faces intersecting V are the coordinate
(n-1)-planes. Exclude points in these coordinate
planes; also exclude points on rays from V to the (n-2)
edges (these are the intersections of adjacent faces).
Both these are sets of zero volume. Any remaining point P
in the cube is in the first quadrant; therefore a ray
through the origin must intersect one of the faces of the
cube disjoint from V. The choice is unique by avoiding
edges.

The volume of the cube of height h is h^n, and the cube
consists of n equal cones over a (n-1) dimensional face,
which is really just an (n-1)-cube. Thus volume of a
height h cone over an (n-1)-cube of side length h is
(h^n)/n. Thus by the linearity, the volume of the cone of
height one over an (n-1)-cube of side length h is
(h^n)/(n*h)=1/n*volume of the (n-1) cube. Since we
think of the sphere as made up of infinitessimal
(n-1)-cubes, the volume of the cone over the unit
sphere=Vol(n)=(1/n)*surface area of the
sphere=A(n-1)/n.

All that is left is to calculate the area of the sphere in n
dimensions=A(n-1). First consider the 3D case for the sphere of radius
one. Since the surface area of the region from phi1 to phi2 in
spherical coordinates is 2*pi*(cos(phi1)-cos(phi2)), which is also
equal to the surface area of the portion of the cylinder of radius one
which surrounds this region (see Figure 4). For a more elementary
proof involving similar triangles, see Figure 5. Now we only need the
surface area of the cylinder. Thus A(2)=2*pi*2. In nD, the
generalization of the statement is true. Using n dimensional spherical
coordinates, the same pair of similar triangles show that the
(n-1)-sphere has the same area as the cross product of the circle with
the (n-2)-ball. That is, A(n-1)=2*pi*Vol(n-2).

So finally, since Vol(n)=1/n*A(n-1), we have
Vol(n)=2*pi/n*Vol(n-2). All these results were for
radius one spheres and balls. To get the volume of the
inscribed ball in a unit cube VI(n), just divide by 2^n. In
other words, VI(n)=pi/(2*n)*VI(n-2). This was a result
quoted in the beginning of the article.














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