swinging through an Arc of Finite Magnitude. 853 



Then by projecting the figure onto the circumscribino- 

 cylinder whose circle of contact is the polar of the point A, 



after the manner of Archimedes, we find that the area of the 

 strip UDD'C is equal to R' 2 cos AC dK. If we take 



R 2 = 



the area of the strip can be written 



cos AC dA 



and thus it appears that the integral will be equal to the 

 area outside the locus of C and inside the polar of the point 

 A, when the angle A proceeds from an initial value zero to a 

 final value (f>. 



We next proceed to find the locus of C. 



Produce the side AC to meet the polar circle of A, and let 

 the point of intersection be called D. In like manner produce 



