# Volume of Cones or Cylinders of Same Height are in Same Ratio as Bases

## Theorem

In the words of Euclid:

Any cones and cylinders which are of the same height are to one another as their bases.

## Proof

Let there be cones and cylinders of the same height.

Let $ABCD$ and $EFGH$ be their bases.

Let $L$ and $M$ be their apices.

Let $AC$ and $EG$ be the diameters of their bases.

It is to be demonstrated that:

$c \left({ABCD}\right) : c \left({EFGH}\right) = AL : EN$

where:

$c \left({ABCD}\right)$ is the circle $ABCD$
$c \left({EFGH}\right)$ is the circle $EFGH$
$AL$ is the cone whose base is $c \left({ABCD}\right)$
$EN$ is the cone whose base is $c \left({EFGH}\right)$.

Suppose it is not the case that $c \left({ABCD}\right) : c \left({EFGH}\right) = AL : EN$.

Then:

$c \left({ABCD}\right) : c \left({EFGH}\right) = AL : O$

where $O$ is some solid figure either less than or greater than the cone $EN$.

Suppose $O$ is less than the cone $EN$.

Let $X$ be a solid figure equal to $EN - O$.

Therefore:

$EN = O + X$

Let the square $\Box EFGH$ be inscribed in $c \left({EFGH}\right)$.

$\Box EFGH$ is greater than half $c \left({EFGH}\right)$.

Let a pyramid be set up on the base $\Box EFGH$ of equal height as the cone.

From:

Proposition $6$ of Book $\text{XII}$: Sizes of Pyramids of Same Height with Polygonal Bases are as Bases

and:

Square Inscribed in Circle is greater than Half Area of Circle

it follows that:

the pyramid so set up is greater than half the cone.

Let the arcs $EF, FG, GH, HE$ be bisected at $P, Q, R, S$.

Let $HP, PE, EQ, QF, FR, RG, GS, SH$ be joined.

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

each of $\triangle HPE, \triangle EQF, \triangle FRG, \triangle GSH$ is greater than half the segment of the circle $EFGH$ around it.

On each of $\triangle HPE, \triangle EQF, \triangle FRG, \triangle GSH$, let pyramids be set up with the same height as the cone.

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

each of these pyramids is greater than half the segment of the cone around it.

The operation:

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

can be repeated indefinitely.

eventually we will leave some segments of the cone which will be less than $X$.

Let such segments be left, and let them be $HP, PE, EQ, QF, FR, RG, GS, SH$.

Therefore the remainder, the pyramid whose base is the polygon $HPEQFRSH$ and whose height equals the height of the cone, is greater than $O$.

Let the polygon $DTAUBVCW$ similar to $HPEQFRSH$ be inscribed in $c \left({ABCD}\right)$.

Let a pyramid be set up on the base $DTAUBVCW$ the same height as the cone $EN$.

$AC^2 : EG^2 = DTAUBVCW : HPEQFRSH$

But we have that:

$c \left({ABCD}\right) : c \left({EFGH}\right) = AL : O$
$DTAUBVCW : HPEQFRSH = DTAUBVCWL : HPEQFRSHN$

where:

$DTAUBVCWL$ is the pyramid whose base is $DTAUBVCW$ and whose apex is $L$
$HPEQFRSHN$ is the pyramid whose base is $HPEQFRSH$ and whose apex is $N$.
$AL : O = DTAUBVCWL : HPEQFRSHN$
$AL : DTAUBVCWL = O : EN$

But $AL > DTAUBVCWL$.

Therefore $O > EN$.

But by hypothesis, $O < EN$.

From this contradiction it follows that it cannot be the case that:

$c \left({ABCD}\right) : c \left({EFGH}\right) = AL : O$

where $O$ is some solid figure less than the cone $EN$.

Similarly it can be proved that it cannot be the case that:

$c \left({EFGH}\right) : c \left({ABCD}\right) = EN : O$

where $O$ is some solid figure less than the cone $AL$.

It remains to be shown that it is not the case that:

$AL : O = c \left({ABCD}\right) : c \left({EFGH}\right)$

where $O$ is a solid greater than $EN$.

Suppose the contrary, that there exists a solid $O$ greater than $EN$ such that:

$AL : O = c \left({ABCD}\right) : c \left({EFGH}\right)$

That is:

$c \left({EFGH}\right) : c \left({ABCD}\right) = O : AL$

But:

$O : AL = EN : X$

where $X$ is some solid less than $AL$.

Therefore:

$c \left({EFGH}\right) : c \left({ABCD}\right) = EN : O$

which was demonstrated to be impossible.

Therefore it is not the case that:

$AL : O = c \left({ABCD}\right) : c \left({EFGH}\right)$

where $O$ is a solid greater than $EN$.

But it was also proved that it is not the case that:

$AL : O = c \left({ABCD}\right) : c \left({EFGH}\right)$

where $O$ is a solid less than $EN$.

Therefore:

$AL : EN = c \left({ABCD}\right) : c \left({EFGH}\right)$
$AL : EN = k \left({ABCD}\right) : k \left({EFGH}\right)$

where $k \left({ABCD}\right)$ and $k \left({EFGH}\right)$ are the cylinders whose bases are $c \left({ABCD}\right)$ and $c \left({EFGH}\right)$ and whose heights are equal.

$\blacksquare$

## Historical Note

This proof is Proposition $11$ of Book $\text{XII}$ of Euclid's The Elements.