ILLUSTRATIVE EXAMPLE 



141 



METHOD III. Geometrical Method. The center of gravity can also be found 

 by regarding the given volume as the sums 

 and differences of simpler volumes of which 

 the center of gravity is already known. 



The volume is obtained by taking the 

 complete sphere OPsQ and subtracting from 

 it the cone OPrQ and the spherical segment 

 PrQs. The center of gravity of the sphere 

 and cone are known, that of the segment 

 PrQs is most easily found by regarding it as 

 the difference between the sector CPsQ and 

 the cone CPrQ. Thus we regard the original 

 figure as made up of -p IG 



(sphere OPsQ) - (cone OPrQ) - (sector CPsQ) + (cone CPrQ). 



The volumes of these, and the distances of their centers of gravity from 

 measured along OC, are as follows : 



FIGURE 



VOLUME 



+ sphere 

 cone OPr Q 

 - sector CPsQ 

 -f cone CPrQ 



f Tra 3 



$ (2 a cos 2 a) (Tra 2 sin 2 2a) 



- f Tra 3 (1 - cos 2a) 



(a cos 2a) (Tra 2 sin 2 2a) 



DISTANCE OF C.G. FROM O 



a 



f(2 a cos 2 a) 



a + | a (1 + cos 2a) 



a + f a cos 2a 



In this table the negative sign denotes that a figure is to be removed, so that 

 its volume must be reckoned as of negative sign. 



Denoting the distance of any center of gravity from O by x, and using 

 the formula 



of 86, we obtain as the distance of the center of gravity of the whole figure 

 from O 



~ 



-|(2acos 2 ar) 2 (Tra 2 sin 2 2ar)-|Tra 4 (l-cos2a:){l+|(l + cos2a:)} 

 | Tra 3 - 1 (2 a cos 2 a) (Tra 2 sin 2 2a) - f Tra 3 (l - cos2a) -f i(acos2a)(Tra 2 sin 2 2a) 



+ |(a cos 2a) (Tra 2 sin 2 2a:)a(l + f a cos 2 a) 

 | Tra 3 - |(2a cos 2 a)(7ra 2 sin 2 2a) - f Tra 3 (l- cos2a)+|(a cos 2 a) (Tra 2 sin 2 2 a) ' 



which, after reduction, gives 



x = a cos 2 or, 

 the same result as before. 



