# Volume of Cone is Third of Cylinder on Same Base and of Same Height

## Contents

## Theorem

In the words of Euclid:

(*The Elements*: Book $\text{XII}$: Proposition $10$)

## Proof 1

Let a cone and a cylinder have the same base $ABCD$.

Let them be of equal height.

Suppose the cylinder is not three times the volume of the cone.

Then the cylinder will either be greater than or less than three times the volume of the cone.

First suppose that the cylinder is greater than three times the volume of the cone.

From Proposition $6$ of Book $\text{IV} $: Inscribing Square in Circle:

From Square Inscribed in Circle is greater than Half Area of Circle:

From the square $ABCD$, let a prism be set up whose height equals the height of the cylinder.

Again using the reasoning of:

in conjunction with

this prism is greater than half the cylinder.

Let the arcs $AB, BC, CD, DA$ be bisected at $E, F, G, H$.

Let $AE, EB, BF, FC, CG, GD, DH, HA$ be joined.

From Proposition $2$ of Book $\text{XII} $: Areas of Circles are as Squares on Diameters:

- each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$ is greater than half the segment of the circle $ABCD$ around it.

On each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$, let prisms be set up with the same height as the cylinder.

Again from the reasoning in Proposition $2$ of Book $\text{XII} $: Areas of Circles are as Squares on Diameters:

The operation:

- bisecting the arcs remaining, joining the points of bisection with straight lines and setting up prisms on the resulting triangles the same height as the cylinder

can be repeated indefinitely.

From Proposition $1$ of Book $\text{X} $: Existence of Fraction of Number Smaller than Given:

- eventually we will leave some segments of the cylinder which will be less than the amount by which the cylinder is greater than triple the cone.

Let such segments be left, and let them be $AE, EB, BF, FC, CG, GD, DH, HA$.

The remainder, the prism whose base is the polygon $AEBFCGDH$ and whose height equals the height of the cylinder, is then greater than three times the cone.

- the prism whose base is the polygon $AEBFCGDH$ and whose height equals the height of the cylinder is three times the pyramid whose base is $AEBFCGDH$ and whose apex is the same as the apex of the cone.

Therefore the pyramid whose base is $AEBFCGDH$ and whose apex is the same as the apex of the cone is greater than the cone which has the circle $ABCD$ as base.

But it is also less, as it is enclosed by it.

From this impossibility it follows that the cylinder cannot be greater than three times the volume of the cone.

Now suppose that the cylinder is less than three times the volume of the cone.

That is, the cone is greater than a third part of the cylinder.

From Proposition $6$ of Book $\text{IV} $: Inscribing Square in Circle:

From Square Inscribed in Circle is greater than Half Area of Circle:

From the square $ABCD$, let a pyramid be set up whose apex is the same as the apex of the cone.

Again using the reasoning of:

in conjunction with the reasoning of:

this pyramid is greater than half the cone.

Let the arcs $AB, BC, CD, DA$ be bisected at $E, F, G, H$.

Let $AE, EB, BF, FC, CG, GD, DH, HA$ be joined.

From Proposition $2$ of Book $\text{XII} $: Areas of Circles are as Squares on Diameters:

- each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$ is greater than half the segment of the circle $ABCD$ around it.

On each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$, let pyramids be set up whose apex is the same as the apex of the cone.

Again from the reasoning in Proposition $2$ of Book $\text{XII} $: Areas of Circles are as Squares on Diameters:

The operation:

- bisecting the arcs remaining, joining the points of bisection with straight lines and setting up pyramids on the resulting triangles whose apex is the same as the apex of the cone

can be repeated indefinitely.

From Proposition $1$ of Book $\text{X} $: Existence of Fraction of Number Smaller than Given:

- eventually we will leave some segments of the cone which will be less than the amount by which the cone is greater than one third the cylinder.

Let such segments be left, and let them be $AE, EB, BF, FC, CG, GD, DH, HA$.

The remainder, the pyramid whose base is the polygon $AEBFCGDH$ and whose apex is the same as the apex of the cone, is then greater than one third the cylinder.

- the pyramid whose base is the polygon $AEBFCGDH$ and whose apex is the same as the apex of the cone is one third the prism whose base is $AEBFCGDH$ and whose height equals the height of the cylinder.

Therefore the prism whose base is $AEBFCGDH$ and whose height equals the height of the cylinder is greater than the cylinder which has the circle $ABCD$ as base.

But it is also less, as it is enclosed by it.

From this impossibility it follows that the cylinder cannot be less than three times the volume of the cone.

The result follows.

$\blacksquare$

## Proof 2

Let the cone be of height $h$.

Let the area of the base of the cone be $A$.

From Volume of Cylinder, the volume of a cylinder of base $A$ and height $h$ is $A h$.

Let the cone be divided by planes parallel to its base each positioned some small distance $d$ apart.

Let $d$ be sufficiently small that they can be approximated to cylinders in shape.

Let there be $n$ of these small cylinders in total.

Let the volumes of each of these small cylinders be:

- $v_1, v_2, v_3, \ldots, v_n$

starting from the base of the cone and working up.

We have that:

- $v_k = d a_k$

where $a_n$ is the

Let the areas of the bases of each of these small cylinders be:

- $a_1, a_2, a_3, \ldots, a_n$

starting from the base of the cone and working up.

## Historical Note

This theorem is Proposition $10$ of Book $\text{XII}$ of Euclid's *The Elements*.

The treatise *The Method* of Archimedes, discovered in $1906$, tells us that this result was discovered by Democritus.