Frequently Asked Questions
What is the volume of a pyramid?
This problem was first stated around 400 BCE by the Greek mathematician and philosopher, Democritus but proved by the brilliant mathematician Eudoxus (408-305 BCE). He reasoned that a pyramid can be approximated to a collection of of slabs of reducing size. For example, a square based pyramid is comparable to pile of square based cuboids.
It is quite clear that this model will not provide the precise volume of a square based pyramid, but it will be a close approximation. Eudoxus argued that by increasing the number of slabs and reducing their heights the approximation would improve. If the slabs are sufficiently thin (infinitesimally so) we will have found the volume of a pyramid.
Let us consider the case of a square based pyramid. From the diagram below it should be evident that the volume of the pyramid is less than one-half of the volume of the box that contains it.
It may be possible for the reader to see that the pyramid occupies over one-quarter of the box, hence its volume lies between one-quarter and one-half of the box in which it is contained. What Eudoxus found, using the calculus, is that any pyramid will occupy precisely one-third of the prism that contains it.
Theorem
The volume of a pyramid, V, is given by V = (A_{base} × h)/3, where h is the perpendicular height.
Proof
Consider any pyramid, perpendicular height h and with area of base A.
Imagine that the pyramid is split into n layers. By similarity the k th layer will have a base with dimensions k/n as a fraction of the original base. So the area of the k th base will be (^{k}/_{n})^{2}A.
The area of the base of each layer will be (^{1}/_{n})^{2}A, (^{2}/_{n})^{2}A, ... , (^{n}/_{n})^{2}A.
By considering each layer to be a prism, each will be h/n units tall and the volume of the k th slab will be (^{h}/_{n}) × (^{k}/_{n})^{2}A.
Hence the volume of the pyramid, V, can be approximated.
V | ≈ (^{h}/_{n})(^{1}/_{n})^{2}A + (^{h}/_{n})(^{2}/_{n})^{2}A + (^{h}/_{n})(^{3}/_{n})^{2}A + ... + (^{h}/_{n})(^{n}/_{n})^{2}A |
= (^{Ah}/_{n3})(1^{2} + 2^{2} + 3^{2} + ... + n^{2}) |
Using the result
1^{2} + 2^{2} + 3^{2} + ... + n^{2} = ^{1}/_{6} n(n + 1)(2n + 1)
We get
V | ≈ (^{Ah}/_{n3}) ^{1}/_{6} n(n + 1)(2n + 1) |
= (^{Ah}/_{6})(1 + ^{1}/_{n})(2 + ^{1}/_{n}) |
Clearly the more layers, i.e. the greater the value of n, the better the approximation. So as n tends towards infinity, we can see that ^{1}/_{n} becomes less significant (tending towards zero).
Hence V tends towards (^{Ah}/_{6})(1 + 0)(2 + 0) = ^{Ah}/_{3}.
So it has been demonstrated that the volume of any pyramid is one-third the volume of the prism that contains it.
Also notice that this method of proof is not dependent on the type of pyramid. All that was needed was the area of the base and the perpendicular height and so it would also apply to the cone.