Unizor - Geometry3D - Pyramids - Volume as Limit
Channel: Zor Shekhtman on UNIZOR Education
Date: 2015-09-09

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Volume of Pyramids as Limit

We strongly recommend to review a lecture that introduces a concept of a pyramid under a topic "Elements of Solid Geometry".
It introduces a pyramid-related terminology we will use in this and other lectures.

We also suggest to review a lecture Area as Limit in the previous topic 3-D Similarity, since we are going to use an analogous method to evaluate the volume of a pyramid, just extend it to a three-dimensional case.

The most practical aspect of a theory of pyramids is their volume. This lecture will study this issue and we will derive a formula for a volume of the pyramid.
We will approach this problem from two different angles, both not absolutely rigorous, but intuitively acceptable. The rigorous proof is based on more advanced topics studied in Calculus.

Approach 1

We will approximate a volume of a pyramid with a volume of an object that consists of little steps around this pyramid and getting close to its slanted side faces.

Assume that SABC is a triangular pyramid with a base plane β that contains triangle ΔABC and apex (top vertex) S.

Drop an altitude SH from apex S onto a base plane β (point H is the base of this perpendicular).
Let's divide segment SH into N equal parts and label the division points (sequentially, from S to H) as H1, H2, ...,HN-1. We will identify point H as HN for convenience.

Draw N-1 planes parallel to base plane β through each division point on altitude SH. The plane #n, that we will call βn, going through point Hn, intersects our pyramid at triangle ΔAnBnCn, where n is any integer number from 1 to N-1. We will identify point A, B and C as AN, BN and CN correspondingly for convenience.

Construct a short right prism using triangle ΔAnBnCn as a bottom base and the plane βn−1 just above it as a plane where the top base is located. Let this top base be triangle ΔA'nB'nC'n.

It's intuitively acceptable (and we are not going to prove it rigorously now because of complexity) that the composition of all these short prisms resembles the shape of a pyramid, that the resemblance is better when the number of prisms increasing, while the height of each decreasing and that the combined volume of these prisms approximates the volume of a pyramid with the approximation becoming more and more precise as the number of prisms N grows to infinity.

So, let's evaluate the combined volume of these prisms and determine the limit it tends to as N→∞.

As we know, the volume of a prism equals to a product of the area of the base by its height.
The height of each prism is the same and equals to 1/N of the height of a pyramid AH.
The area of the base of the prism #n is the area of the triangle ΔAnBnCn. To evaluate it, consider similarity between this triangle and the base of a pyramid - triangle ΔABC. The similarity is very easy to prove based on a scaling with a center at the apex S and the factor n/N. As we know (see 3D Similarity topic), similar flat objects have their areas proportional to a square of the scaling factor. Therefore, the area of triangle ΔAnBnCn equals to the area of triangle ΔABC multiplied by a factor n²/N².

Let the height of our pyramid AH be equal to h and the area of its base triangle ΔABC be equal to s. Then the prism #n has volume equal to
vn = (s·n²/N²)·(h/N)
Simplifying this, we get
vn = (s·h/N³)·n²

The next step is to summarize this expression for all n from 1 to N.
Since s, h and N are constants, Σn∈[1,N](vn) =
= (s·h/N³)·Σn∈[1,N](n²)

It's easy to derive that Σ[n²] = N(N+1)(2N+1)/6

Using this in the formula for a volume of an object that contains short prisms stacked on the top of each other, we obtain the following:
Σn∈[1,N](vn) =
= (s·h/N³)·N(N+1)(2N+1)/6 =
= s·h/3+s·h·/2N+s·h·/6N²

As N→∞, the above expression tends to s·h/3, which is the formula for a volume of a pyramid:
A PRODUCT OF THE AREA OF A BASE AND ONE THIRD OF THE ALTITUDE.