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) = πHI×HK2. 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 = πHK. With some algebraic simplification, the proportional version of the Pappus-Guldin theorem from the last section then becomes
\[\eqalign{ {rev(AB) \over \pi HI\times HK^2 } &= {area(AB) \over HI\times HK} \times { circum(a) \over \pi HK} \cr &= {area(AB) \times circum(a) \over \pi HI \times HK^2} \cr }\]
In particular, the denominators on both sides of the equation are the same. Consequently, the numerators must be equal as well. That is,
\[rev(AB) = area(AB) \times circum(a)\]
which is precisely the Pappus-Guldin theorem.