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James Gregory and the Pappus-Guldin Theorem - Gregory's Proof Revealed

Author(s): 
Andrew Leahy (Knox College)

 

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.

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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.

Dummy View - NOT TO BE DELETED