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## The Volume of a Hyperpyramid

On this page, we will derive the formula for the volume of a regular n-dimensional hyperpyramid -- that is to say, a hyperpyramid whose base is an (n-1)-dimensional hypercube with edge length r, and whose height is r.  We'll show that the volume of such an n-pyramid 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 n-pyramid, or n dimensional hyperpyramid, is analogous to a triangle in arbitrary dimensions.  It's formed by starting with an (n-1)-cube, and sweeping the cube perpendicular to itself, just as we did in forming an n cube (see hypercubes for a discussion of n-cubes and their volumes).  However, as we sweep it, we shrink it, linearly, down to a point.  We show this for a 3-pyramid 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 n-pyramid, we might guess that the n-volume would be the (n-1)-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 n-cube with n-pyramids, 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 "(n-1) 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 n-pyramid is indeed linear in the height and the n-1 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 n-pyramids of height r, and with r sized (n-1)-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 n-pyramid 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 (n-1)-cube, and growing linearly until, when it's moved r units along the axis, it will have grown into an (n-1)-cube with edges of length r.  This will hopefully become clearer momentarily, as we give some examples!

### 1-Pyramids ... aka Line Segments

For a 1-pyramid, this operation just produces a line segment.  The "1-volume" of this "1-pyramid" is r units, which is, in fact, the same as the volume of an r unit 1-cube ... no surprise, since the "1 pyramid" and the "1 cube" are both just line segments.  So the volume of a regular 1-pyramid is: ### 2-Pyramids ... aka Triangles

 Figure 3: Two triangles in square For a 2-pyramid, 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 "2-volume" is ### 3-Pyramids ... 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 onthe 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 zero-sized 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 rxr square.  It will sweep out a pyramid, r units high, with an rxr 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 3-cube -- 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 3-pyramid, 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 n-Pyramids

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 zero-dimension (n-1)-cube at the origin, and slide it along the x axis, growing it as it goes, we can "see" that we will produce an n-pyramid 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 n-cube with exactly n hyperpyramids, and the volume of a regular n dimensional hyperpyramid must be ### An Alternative Geometric Argument

 Figure 7: Square filled withfour half-height 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 n-1 cube.  Sweep the face exactly perpendicular to itself, directly through the center of the n-cube, 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 2n pyramids, each half the height of the cube, or r/2 units.  Since these half-height 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 3-cube 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 (n-1)-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 (n-1)-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 (x1...xn).  If any two of the xi 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 n-volume, as the faces are (n-1) 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 xi are all different, then find the largest one; call it xk.  Then every coordinate must lie between 0 and xk:

(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 n-pyramid is 1/n times the volume of an n-cube, 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 (n-1)-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 "sub-pyramids" 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 sub-pyramids, 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 "cone-like" 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