For most mathematicians, the ratio that Gregory establishes between the trunk and the solid of revolution is easier to grasp than the ratio he establishes between the trunk and the cylinder. This stems largely from differences in mathematical upbringing. As noted previously, centers of gravity were an important and familiar part of the mathematical culture during the sixteenth and seventeenth centuries. But apart from the intuitive definition of the center of gravity as the balancing point for a region, they are largely absent from the training of most mathematicians today. However, the most important result for an intuitive understanding of centers of gravity is probably familiar to those who have seen second semester calculus:

Suppose "two masses *m*_{1} and *m*_{2} are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances *d*_{1} and *d*_{2} from the fulcrum. The rod will balance if *m*_{1}*d*_{1} = *m*_{2}*d*_{2}" [12, p. 554]. This is a restatement of something first given as Proposition I.6 of Archimedes' *On the Equilibrium of Planes*. After some division, it states the result (more amenable to someone versed in proportion theory) that two volumes balance at a point such that the ratio of the distances of this point to the centers of gravity of the volumes is reciprocally proportional to the ratios of the two volumes. That is, *m*_{1}/*m*_{2} = *d*_{2} / *d*_{1}. In what follows we will occasionally have need to rely on this and other intuitive results on centers of gravity.

Gregory's first task is to locate the center of gravity of the trunk. To start things off, Gregory limits himself to right cylinders constructed from figures that are symmetric about an axis perpendicular to the axis of rotation of the solid of revolution.

This restriction is very useful: If a figure *AB* is symmetric, its center of gravity will lie somewhere on its axis of symmetry. Furthermore, if *a* is the center of gravity of *AB* and *b* is the center of gravity of the top of the right cylindrical figure built on *AB*, then the center of gravity of this entire cylindrical figure lies at the midpoint *X* of the segment *ab*.

In particular, let *IFQS* be a rectangle which (1) passes through the axis of symmetry *IS*, (2) is perpendicular to the axis of rotation, and (3) has the same height *FI = SQ* as the cylindrical figure. Furthermore, suppose *R* and *H* are the respective midpoints of *QS* and *FI*. Then the center of gravity *X* of the cylindrical figure will lie at the intersection of the line segments *ab* and *RH*.

All of this can be used to determine the center of gravity of the trunk. Suppose that the (symmetrical) cylindrical figure is sliced by a plane through the axis of rotation to form an upper trunk and a lower trunk as before. For simplicity of notation, assume that the axis of symmetry *IS* is also a radius of rotation. Reasoning by symmetry again shows that the centers of gravity of both the upper trunk and the lower trunk will lie on the plane *IFQS*.

Since *FI* is twice *HI*, symmetry and Euclid VI.2 together show (with some work) that if the lower trunk is sliced by a plane perpendicular to *IFQS* to form a rectangular slice *KLMN* of the trunk, then *KLMN* will have its geometrical center on the line *HS*.

Since the geometrical center of a rectangle is also its center of gravity, it follows that the center of gravity of such a slice of the trunk will lie on the line *HS*. Since this is true for each slice *KLMN* of the trunk an argument from indivisibles then shows that the center of gravity *Y* of the lower trunk must lie on the line *HS*. A similar argument shows that the center of gravity *Z* of the upper trunk must lie on the line *FR*.

But where on *HS* will the lower trunk's center of gravity lie? As Archimedes had demonstrated, if the mass of the upper trunk is concentrated at its center of gravity *Z* and the mass of the lower trunk is concentrated at its center of gravity *Y*, then the center of gravity of the upper and lower trunk combined will be a point on the segment *YZ* such that the ratio of the distance from this point to *Y* and *Z*, respectively, will be inversely proportional to the ratio of the volume of the upper trunk to the lower trunk. But the two trunks combined make up the entire cylinder, the center of gravity of which is *X*.

It follows that *X*, *Y*, and *Z* are collinear and

{YX \over ZX } = {volume(upper) \over volume(lower) }

where

But *X* lies on *HR*, and *Z* and *Y* lie on *FR* and *HS*, respectively. Also, the lines *HR*, *FQ*, and *IS* are parallel to each other, as are the lines *QS*, *FI*, and *ab*, as well as the lines *FR* and *HS*. So Δ*HYX* is similar to Δ*RZX* and

{Ia \over a S } = {HX \over XR} = { YX \over ZX} = {volume(upper) \over volume(lower)}

The final trick to determining the ratio between the trunk and the cylinder is to use Euclid V.18 (a.k.a. adding "one" to both sides of the above proportion) to arrive at

\eqalign{ { IS \over a S} & = {(Ia + a S) \over a S } \cr &= {Ia \over a S } + 1 \cr &= { volume(upper) \over volume(lower)} + 1 \cr &= {volume(upper) + volume(lower) \over volume(lower)} \cr &= { cyl(AB) \over trunk(AB)}. \cr }

where { trunk(AB) \over cyl(AB) } = {aS \over IS} = {2 \pi aS \over 2 \pi IS} = { circum(a) \over circum(AB) }

where