Commentary on Proposition XII.10

Back to Proposition XII.10.

We observe the use in this proposition of the term “parallelepipedal prism,” which recalls Heron’s “parallelogrammic” or “parallel-sided prism.”

The course of the proof is exactly the same as in XII. 2, except that an arithmetical fraction takes the place of a ratio which, being incommensurable, could only be expressed as a ratio. Consequently we do not need proportions in this proposition, as we did in XII. 2, and shall again in XII. 11, etc.

Euclid exhausts the cylinder and cone respectively by setting up prisms and pyramids of the same height on the successive regular polygons inscribed in the circle which is the common base, viz. the square, the regular polygon of 8 sides, that of 16 sides, etc.

If AB be the side of one polygon, we obtain two sides of the next by bisecting the arc ACB and joining AC, CB. Draw the tangent DE at C and complete the parallelogram ABED.

D C E A B

Now suppose a prism erected on the polygon of which AB is a side, and of the same height as that of the cylinder.

To obtain the prism of the same height on the next polygon we add all the triangular prisms of the same height on the bases ACB and the rest.

Now the prism on ACB is half the prism of the same height on the $\square$ ABED as base. [cf. XI. 28] And the prism on $\square$ ABED includes, and is greater than, the portion of the cylinder standing on the segment ACB of the circle.

The same thing is true in regard to the other sides of the polygon of which AB is one side.

Thus the process begins with a prism on the square inscribed in the circle, which is more than half the cylinder, the next prism (with eight lateral faces) takes away more than half the remainder, and so on; hence [X. 1], if we proceed far enough, we shall ultimately arrive at a prism leaving over portions of the cylinder together less than any assigned volume.

The construction of pyramids on the successive polygons exhausts the cone in exactly the same way.

Now, if the cone is not equal to one-third of the cylinder, it must be either greater or less.

  1. Suppose, if possible, that, $V$, $O$ being their volumes respectively,

$$O > 3V.$$

Construct successive inscribed polygons in the bases and prisms on them until we arrive at a prism $P$ leaving over portions of the cylinder together less than $(O - 3V)$, i.e. such that

$$O > P > 3V.$$

But $P$ is triple of the pyramid on the same base and of the same height; and this pyramid is included by, and is therefore less than, $V$; therefore

$$P < 3V.$$

But, by construction,

$$P > 3V\text{:}$$

which is impossible.

Therefore

$$O \ngtr 3V.$$

  1. Suppose, if possible, that $O < 3V$. Therefore

$$V > \tfrac{1}{3}O.$$

Construct successive pyramids in the cone in the manner described until we arrive at a pyramid $\Pi$ leaving over portions of the cone together less than $(V - \tfrac{1}{3}O)$, i.e. such that

$$V > \Pi > \tfrac{1}{3}O.$$

Now $\Pi$ is one-third of the prism on the same base and of the same height; and this prism is included by, and is therefore less than, the cylinder; therefore

$$\Pi < \tfrac{1}{3}O.$$

But, by construction,

$$\Pi > \tfrac{1}{3}O\text{:}$$

which is impossible.

Therefore $O$ is neither greater nor less than $3V$, so that

$$O = 3V.$$

It will be observed that here, as in XII. 2, Euclid always exhausts the solid by (as it were) building up to it from inside. Hence the solid to be exhausted must, with him, be supposed greater than the solid to which it is to be proved equal; and this is the reason why, in the second part, the initial supposition is turned round.

In this case too Euclid might have approximated to the cone and cylinder by circumscribing successive pyramids and prisms in the way shown, after Archimedes, in the note on XII. 2.