Pyramids are solid figures that have a polygonal base which is joined,
by triangles to a point. A cone is special case of a pyramid where the
polygonal base has become a circle, or an ellipse. Generally, the rules of pyramids
apply to all figures which have all their cross-sectional areas similar and end in a point. A smooth pyramid is one in which the
sections through the point, the centre of the base and the periphery of
the base are triangles. The sides of a smooth pyramid are usually
triangles. Discrete pyramids do not have triangular sections through
their point, centre of the base and their periphery, but are composed
of blocks, and so they have a stepped surface.
In a smooth pyramid, a line through the point and the centre of the
base makes a right-angle.
The volume, V, of a smooth pyramid is:
Where A is the area of the base and h is the vertical height.
In general, the volume of a discrete pyramid is:
where, V is the volume, A is the area of the base, h is the height and
n is the number of layers.
If the base of the pyramid is a triangle, with base, b, and height p,
then the area is 1/2(bp), and the volume is 1/2(bp).h/3:
If the base is a rectangle, a by b, then the area of the base, A=ab,
and the volume is therefore:
A circular cone has a base area of πr2,
so its volume is:
Proof using discrete pyramids
It seems there should be something easy about proving the above, but it
seems there isn't an easy way. It can be demonstrated by making a paper
cone and a cylinder with the same height and base, and filling the cone
with rice to see how many conefuls go into the cylinder. Another way is
to dangle a cone in water and measure the amount of water displaced. In
ancient times, craftsmen must have done this (perhaps when making
things of gold or bronze) and noticed how you seem to get three cones
for every cylinder-full of liquid.
Our first approach will be to consider a discrete pyramid, that is one
made up of blocks. Clearly, as the blocks become infinitesimally small,
the volume of the pyramid becomes the volume of a smooth pyramid.
The following diagram is a vertical section through the middle of a
pyramid (Through the apex and at right angles to the plane of the base):
The theory is that we approximate the volume of the pyramid with the
volume of a corresponding prism. As an illustration, the pyramid is
divided into four. This first part, dividing by 4 is not essential to
the proof. It is just to set the scene for the real proof.
Each of the areas, A1, A2, A3, A4 etc. correspond to the heights h/4,
2h/2, 3h/4 and 4h/4.
The approximate volume of the smooth pyramid (but the volume of the discrete pyramid) is:
The areas, A1..A4 are proportional to the heights
squared, because they are similar figures. So:
So, cancelling out h2
As the height of the prism (used to approximate the volume of the pyramid) is h/4, the volume, V1, is:
and A=A(42/4²), and the corresponding volumes, V1..V2 are each of the areas times the height, h/4.
The volume of the pyramid, approximated as prisms, is the combined volumes:
Where A is the area of the base of the pyramid and V is the volume of the pyramid, approximated as prisms.
Now this is a rough estimate of the volume of a smooth pyramid (but
accurate for a discrete pyramid). However,
had we taken more sections, the volume would have been closer to that of the smooth pyramid.
To do this, we generalise the formula for n sections.
Let us re-write the formula substituting n for
the number of sections (replacing the 4's that refer to the number of sections
If we let n get very large and approach infinity, V will approach the
value of the volume of the pyramid. To find the formula for a smooth pyramid, we need to compute the
sum of the squares of the integers from 1 to n. The sum of the first
n squares of the natural
numbers is given by:
By substituting Equation 2 for the sum of the squares of the natural
numbers in Equation1 and simplifying the n's, we get:
This is the volume of any discrete pyramid with base area, A, and n layers, composed of cubic bricks of side 1.
For instance, if we consider a pyramid composed of cubic
bricks of side 1, making a base area of 9 (A=9), and we take 3 levels
(n=3), the height of the pyramid will also be 3 (n=3), then the volume
of the pyramid will be (or number of bricks required is), substituting values in Equation 3:
9*3(1/3+1/(2*3)+1/(6*32)=14 (units of volume, or cubic bricks of side 1)
of a smooth pyramid
Returning to our smooth pyramid, we can find
its volume by increasing the value of n in Equation
3 to infinity, when the second and
third terms disappear and the volume becomes:
This is the volume, V, for any smooth pyramid with a base Area A.
Proof by calculus
On the left, ACB is a section of a pyramid. AC is a line through the
base, and represents one dimension of the area, A, of the base. DE is any
section through the pyramid at height y, and is parallel to the base, and represents an area Ai,
h is the height of the pyramid.
Because the area section at DE, Ai, is proportional to the area
of the base of the pyramid, A(at AC)
and these areas, A and Ai, are proportional to their heights squared, then:
So, Ai is:
As each component of the total volume is: