Commentary on Proposition XII.11¶
Back to Proposition XII.11.
We need not again repeat the preliminary construction of successive pyramids and prisms exhausting the cones and cylinders.
Let $Z, Z'$ be the volumes of the two cones, $\beta, \beta'$ their respective bases.
If
$$\beta : \beta' \ne Z : Z',$$
then must
$$\beta : \beta' = Z : O,$$
where $O$ is either less or greater than $Z'$.
Suppose, if possible, that $O$ is less than $Z'$.
Inscribe in $Z'$ a pyramid ($\Pi'$) leaving over portions of it together less than $(Z' - O)$, i.e. such that
$$Z' > \Pi' > O.$$
Inscribe in $Z$ a pyramid $\Pi$ on a polygon inscribed in the circular base of $Z$ similar to the polygon which is the base of $\Pi'$.
Now, if $d, d'$ be the diameters of the bases,
$$\begin{aligned} \beta : \beta' &= d^2 : {d'}^2 & \text{[XII. 2]} \\ &= (\text{polygon in } \beta) : (\text{polygon in } \beta') & \text{[XII. 1]} \\ &= \Pi : \Pi'. & \text{[XII. 6]} \end{aligned}$$
Therefore
$$Z : O = \Pi : \Pi',$$
and, alternately,
$$Z : \Pi = O : \Pi'.$$
But $Z > \Pi$, since it includes it; therefore
$$O > \Pi'.$$
But, by construction,
$$O < \Pi' :$$
which is impossible.
Therefore
$$O \nless Z'.$$
Suppose, if possible, that
$$\beta : \beta' = Z : O,$$
where $O$ is greater than $Z'$.
Therefore
$$\beta : \beta' = O' : Z',$$
where $O'$ is some solid less than $Z$.
That is,
$$\beta' : \beta = Z' : O',$$
where $O' < Z$.
This is proved impossible exactly in the same way as the assumption in Part I. was proved impossible.
Therefore $Z$ has not either to a less solid than $Z'$ or to a greater solid than $Z'$ the ratio of $\beta$ to $\beta'$ ; therefore
$$\beta : \beta' = Z : Z'.$$
The same is true of the cylinders which are equal to $3Z, 3Z'$ respectively.