With all of this proportion theory in hand, Gregory's proof of the Pappus-Guldin Theorem falls into place relatively easily. Suppose that *AB* is the geometrical figure which is to be rotated around an axis and that *a* is its center of gravity. The central idea of his proof is to use the proportional version of the theorem given in the last section to compare *AB* with another, easy-to-understand 2-dimensional figure. In this case, that figure is a rectangle *HIJK* and its axis of rotation is simply the side *HI* of the rectangle.

For a rectangular figure *HIJK*, we have *area*(*HIJK*) = *HI*×*HK*. Since the solid of revolution obtained by revolving *HIJK* around the the line *HI* is a cylinder with height *HI* and radius *HK*, we get *rev*(*HIJK*) = π*HI*×*HK*^{2}. Finally, the center of gravity *h* of a rectangle is the geometrical center of the rectangle, so the distance from *h* to *HI* is (1/2) *HK* and thus *circum*(*h*) = 2π×(1/2)*HK* = π*HK*. With some algebraic simplification, the proportional version of the Pappus-Guldin theorem from the last section then becomes

\eqalign{ {rev(AB) \over \pi HI\times HK^2 } &= {area(AB) \over HI\times HK} \times { circum(a) \over \pi HK} \cr &= {area(AB) \times circum(a) \over \pi HI \times HK^2} \cr }

rev(AB) = area(AB) \times circum(a)