On this page, we will derive the formula for the volume of a regular
ndimensional
hyperpyramid  that is to say, a hyperpyramid whose base is an (
n1)dimensional
hypercube with edge length
r,
and whose height is
r.
We'll show that the volume of such an
npyramid
is
(1) 
This is an example of something we could find easily using calculus.
However, in this case we plan to use it to
develop
calculus, so we need to find it some other way. We're going
to
demonstrate this first with a purely geometric argument, using
pictures, and then (since that proof is not quite airtight) we'll prove
it symbolically, down near the end of the page.
Figure
1: Growing a Pyramid:

An
npyramid, or
n
dimensional hyperpyramid, is analogous to a triangle in arbitrary
dimensions. It's formed by starting with an (n1)cube, and
sweeping
the cube perpendicular to itself, just as we did in forming an
n
cube (see
hypercubes
for a discussion of
ncubes and their volumes).
However, as we sweep it, we
shrink it,
linearly, down to a point. We show this for a 3pyramid in
figure 1: We start with a square
and sweep it up, while shrinking it, to form the pyramid.
Though we haven't drawn the process for any other
dimensionality, it's not hard to imagine.
What is the
volume of the pyramid? For a
triangle, it's (1/2) base * height. For an arbitrary
npyramid,
we might guess that the
nvolume would be the
(n1)volume of the base,
times the height,
times
some constant. We need to confirm this guess, and we
need to
determine what the constant might be. We'll do this by
packing an
ncube with
npyramids,
and see how many it takes to exactly fill it up.
Figure
2: A skewed triangle

Before we proceed with finding the volume, we need a simple
observation: If
we
skew a pyramid (or any simple solid), while
keeping the same cross sectional thickness, it doesn't change in
volume. At risk of belaboring the obivous, we'll talk at some
length about this.
We can see it by imagining we slice the pyramid
into a stack of plates. The volume of the pyramid is the sum
of the volumes of the plates.
Skewing the
pyramid is just sliding the plates across each other; it doesn't change
the volume of the individual plates, and doesn't change their number,
and hence doesn't change the net volume of the pyramid. This
is shown for a triangle in
figure 2.
Since the "thickness" of a pyramid scales down
linearly
from the base to the peak, each plate is shaped exactly like the base
(but scaled down). The scale factor for the plates varies
from 1
(at the base) to 0 (at the peak), and depends on how far up the pyramid
the plate is  but not on whether it's been pushed off to one side.
Of course, in our
figure 2,
the plates are rather thick, and I cheated by showing them with beveled
edges; to make a smooth triangle from plates with "squared off"
edges,
you'd need to make the plates very thin  infinitesimally thin.
Let's consider what happens if we
stretch the
pyramid vertically. We can change its height by a factor of
k
just by changing the
thickness of each
plate
by a factor if
k. That
obviously increases the volume of each plate by a factor of
k,
and since the pyramid's volume is the sum of the plate volumes, it also
changes the pyramid's volume by a factor of
k.
Similarly, changing the area (or "(
n1)
volume") of the base by a factor of
s
would in turn scale the volume of each plate by a factor of
s
as well, and so in turn scale the volume of the pyramid by
s
as well. So the volume of an
npyramid is
indeed linear in the height and the
n1 volume of
the base, and so it must just be a constant times those values.
We still need to find the constant.
We have determined that all
npyramids of height
r,
and with
r sized (
n1)cubes
for their bases, must have the same volume, so we can choose any
"shape" of pyramid to examine. For our purposes, we will be
using
right pyramids: The apex is located
directly over one
vertex of the base, and one of the edges leading to the apex intersects
the base at a right angle.
We will form a right
npyramid
on its side,
by starting with a point at the origin. We'll send the point
along the
x axis, and as it moves along the axis it
will expand, becoming an (
n1)cube, and growing
linearly until, when it's moved
r
units along the axis, it will have grown into an (
n1)cube
with edges of length
r.
This will hopefully become clearer momentarily, as we give some examples!
1Pyramids ... aka Line Segments
For a
1pyramid, this operation just produces a
line segment. The "1volume" of this "1pyramid" is
r
units, which is, in fact, the same as the volume of an
r
unit
1cube ... no surprise, since the "1 pyramid"
and the "1 cube" are both just line segments. So the volume
of a regular 1pyramid is:
2Pyramids ... aka Triangles
Figure
3: Two triangles in square

For a
2pyramid,
this produces a right triangle, and when we try to draw the result, we
realize that the boundary of the triangle traveled on a straight line
across a square. The operation has in fact left us exactly
enough
space to do it a
second time, this time "growing" a
triangle along the
y axis. (See
figure 3). We can fit
two
triangles into a square. Consequently, we have just confirmed
what we already knew, which is that
and more specifically, in the specific case we're concerned
with, the base is
r
units wide and the height is
r
units, and the "2volume" is
3Pyramids ... Ordinary "Pyramids"
Figure
4a:
Cube, to be filled
with pyramids:

We now enter more interesting territory. Before writing this
page I didn't know you could do what we're about to do: We're
going to pack three pyramids into a cube, exactly filling it.
Figure
4b:
A pyramid on the X axis:

Let's start by building a pyramid along the
x
axis, must the way we built a triangle on the
x
axis in the previous section: We'll start with a point at the
origin, and slide it
r units to
the right. Actually our "point" is a zerosized
square.
As it moves to the right, we'll
grow it,
until, when it has moved
r units
along the
x axis it will be an
rx
r
square. It will sweep out a pyramid,
r
units high, with an
rx
r
base. This is shown in
figure
4b. Note that the pyramid is
lying on its
side. Note also that we didn't fill in the bottom,
as it seemed easier to understand the picture with it drawn this way.
Now we need to realize something about hypercubes: They're
symmetric
with respect to the axes  interchanging axes doesn't change the
apparent shape of the cube. A hypercube  or an ordinary 3cube
 has exactly the same structure along each axis. Consequently,
it really didn't make any difference which axes we "grew" our pyramid
along; we could insert it in the cube regardless.
In figures
5a through
5c we've shown the pyramid lying on the
x axis, inserted in the cube shown in
figure 4a, and we've shown the equivalent pyramids along the
y and
z axes, also embedded in the cube.
Figure
5a:
Pyramid on X, in cube:

Figure
5b:
Pyramid on Y, in cube:

Figure
5c:
Pyramid on Z, in cube:

In
figure 6a we show the
x and
y pyramids both inserted in the cube, and finally in
figure 6b
we show all three pyramids in the cube. Examining the faces of
the pyramids which lie inside the cube, and noting how they fit
together, we can see that they must, indeed, exactly fill the cube,
just as they are drawn.
And so we see that the volume of a regular 3pyramid, with a base
r units square, and with height
r, must be
Note, when looking at figures (6), that the apexes of all three pyramids are at the
origin.
Unfortunately perspective drawings can be deceiving and it's
rather easy to "see" the common apex in figure (6b) as lying in the
center of the cube!
Figure
6a:
Pyramids on X and Y:

Figure
6b:
3 Pyramids in a Cube:
views

General nPyramids
Again, we need to keep in mind the fact that hypercubes are symmetric with respect to any interchange of the axes.
With that in mind, if we imagine we start with a zerodimension (
n1)cube at the origin, and slide it along the
x axis, growing it as it goes, we can "see" that we will produce an
npyramid lying on the
x axis, just as we produced first a triangle and then a square along the
x
axis. We can also see ... perhaps dimly ... that we can do the
same exact operation along each axis, and it seems clear that the faces
must "fit together" as we perform this operation.
Consequently we can fill an
ncube with exactly
n hyperpyramids, and the volume of a regular
n dimensional hyperpyramid must be
An Alternative Geometric Argument
Figure 7: Square filled with four halfheight triangles

We can present a slightly more complex argument which may be easier to picture.
Instead of starting with a point at the origin, let's start with an
n cube,
r units on an edge. Pick one of its faces  which is, itself, an
n1 cube. Sweep the face exactly perpendicular to itself, directly through the center of the
ncube, and on to the
opposite face. But here's the trick:
shrink it as it moves  shrink it
twice as fast as it moves, so that when it's moved
r/2 units (and has just reached the center of the
n cube), it has shrunk down to nothing. And then grow it again, so that when it's moved a total of
r units, and has reached the opposite face, it has again grown back to its starting size. We can do this
again, once for each axis, and since the faces each shrink linearly to points as they move to the center, they won't "collide".
This
is shown for a square, in
figure 7. We see that we've filled the
square with four triangles  two along each axis. In general, it
should be clear that this operation will exactly fill an
n cube with 2
n pyramids, each
half the height of the cube, or
r/2 units. Since these halfheight pyramids each have half the volume of a pyramid
r units high, we again see that the volume of a pyramid
r units high with a base
r units on an edge must be
It's helpful in picturing this to draw a
3cube
cut up into 6 pyramids. Unfortunately, while I've sketched such a
picture it still needs considerable work before it's ready to put on
this page.
A Symbolic Proof
The above arguments can best be characterized as "motivating" the result. Because we can't actually draw what happens in
n dimensions they don't constitute a solid proof. So, we'll now present a brief algebraic proof.
We know that the vertices of an
n cube,
r units on an edge, consist of all points for which every coordinate is either
0 or
r:
(p.1)
The region
contained in the
n cube consists of all points which lie within
r units of the origin along each axis:
(p.2)
A
pyramid lying on the
k axis is the region swept out by an (
n1)cube as it moves along axis
k,
growing as it goes. At a point
s units from the origin on the
k axis, the cube will be
s units on a side. Consequently, at that point, all coordinates of all points in that (
n1)cube must fall within
s units of 0. This may sound a little like a we're describing a sphere, but we're not  we're bounding the value of
each coordinate, not the total distance from the origin. The set of points contained in such a pyramid must be:
(p.3)
Pick
any arbitrary point within the
n cube. Call it (x
_{1}...x
_{n}). If any two of the x
_{i} are equal then the point lies on a
face (a face of the
n cube, or a face of a pyramid contained in it),
and the union of all faces has zero
nvolume, as the faces are (
n1)
dimensional objects; consequently we can ignore them (they don't affect the overall volume). To put it another way, in
n+1 dimensional space, the
n dimensional faces are "flat"  they contain no volume.
If, on the other hand, the values of the x
_{i} are all
different, then find
the largest one; call it x
_{k}. Then every coordinate must lie between 0 and x
_{k}:
(p.4)
And so the point does lie in the pyramid along axis
k. Furthermore,
this is true for
no other coordinate (it's only true for the coordinate with the
largest value), so the point does not lie in any
other pyramid.
Therefore each point which does not lie on a face lies in
exactly one of the pyramids.
There are
n axes, there are
n pyramids, so the volume of an
npyramid
is 1/n times the volume of an
ncube, which was to be shown.
A Small Generalization
We
have already observed that the angle the vertical axis of the
pyramid makes with the base doesn't make any difference to its volume  it's just the
(hyper)area of the base, and the overall height, which matter.
What's more, the
shape
of the base makes no difference, either. If we think about the
"stack of plates" metaphor, we can see that the shape of each plate
doesn't matter; the plates scale down in volume the same way as we go
up the pyramid no matter how they're shaped. Thus, the volume of
the object should go as 1/n times the height times the (
n1)volume of the base, no matter what shape the base has.
Alternatively,
we can divide the base plate of an arbitrary pyramid into tiny
(hyper)squares, and hence divide the volume of the whole pyramid into
pyramids with "square" bases. Since each of those "subpyramids"
will have volume 1/n times the (hyper)area of its base times its
height, and the volume of the whole object must be the sum of the
volumes of the subpyramids, the volume of the whole must again be 1/n
times the area of its base times its height.
In three
dimensions, this implies cones, tetrahedrons, and other "conelike"
shapes will all have volume equal to the area of the base times 1/3 the
height.
Page
created on 10/06/2007; first posted on 11/04/2007