P
# Ken Ward's Mathematics
Pages

## Pyramids
(Volumes)

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.

### Triangular
Base

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:

### Rectangular
Base

If the base is a rectangle, a by b, then the area of the base, A=ab,
and the volume is therefore:

### Cone

A circular cone has a base area of πr^{2},
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.

### Discrete
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, A_{1}..A_{4} are proportional to the heights
squared, because they are similar figures. So:

So, cancelling out h^{2}

As the height of the prism (used to approximate the volume of the pyramid) is h/4, the volume, V_{1}, is:

Similarly, A_{2}=A(2^{2}/4²),
A_{3}=A(3^{2}/4²),
and A=A(4^{2}/4²), and the corresponding volumes, V_{1}..V_{2} are each of the areas times the height, h/4.

The volume of the pyramid, approximated as prisms, is the combined volumes:

Or simplifying:

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

[1]

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:

[2]

By substituting Equation 2 for the sum of the squares of the natural numbers in Equation1 and simplifying the n's, we get:

[3]

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*3^{2})=14 (units of volume, or cubic bricks of side 1)

### Volume
of a smooth pyramid

This is the volume, V, for any smooth pyramid with a base Area A.

## Proof by calculus

_{i},

h is the height of the pyramid.

_{i}, is proportional to the area
of the base of the pyramid, A (at AC)
and these areas, A and A_{i}, are proportional to their heights squared, then:

So, A_{i} is:

As each component of the total volume is:

Substituting the formula for the area:

We now have the means of integrating to get the volume, V, between the limits 0 and h:

And by integrating:

Or:

Ken Ward's Mathematics Pages

# Faster Arithmetic - by Ken Ward

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.

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 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, A

So, cancelling out h

As the height of the prism (used to approximate the volume of the pyramid) is h/4, the volume, V

Similarly, A

The volume of the pyramid, approximated as prisms, is the combined volumes:

Or simplifying:

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

[1]

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:

[2]

By substituting Equation 2 for the sum of the squares of the natural numbers in Equation1 and simplifying the n's, we get:

[3]

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*3

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.

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 A

h is the height of the pyramid.

Because the area section at DE, A

So, A

As each component of the total volume is:

Substituting the formula for the area:

We now have the means of integrating to get the volume, V, between the limits 0 and h:

And by integrating:

Or:

Ken Ward's Mathematics Pages

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: