140 



CENTER OF GRAVITY 



Substituting these values, we find 



r = 



cos 6 a. 



= a cos 2 a. 



Thus the center of gravity is on the axis of the cone at a distance a cos 2 a 

 from the vertex. 



METHOD II. Cartesian coordinates. We may next employ Cartesian coordi- 

 nates, taking as origin and the axis of the cone as axis of x. The equation of 



the cone is now 



while that of the sphere is 



x 2 + y 2 + z 2 - 2 ax = 0. 

 From 106, we have 



= 



In each integral we may inte- 

 grate first with respect to y and z 

 together. We have to evaluate 

 the same integral in both cases, namely \ Cdydz, the limits being given by 



FIG. 82 



and y 2 + z 2 = 2 ax x 2 . 



The problem is the same as that of finding the area of a circular ring of 

 inner and outer radii xtano: and V2 ax x 2 respectively. (This ring is, of 

 course, the intercept of the solid on the plane parallel to the yz plane.) The 

 area of the ring is 



IT (2 ax x 2 ) - TT (a; 2 tan 2 a) = TT (2 ax - x 2 sec 2 a), 



and on substituting this value for J \ dydz, the formula becomes 



TTX (2 ax x 2 sec 2 a) dx 







AT (2 ax - x 2 sec 2 a) dx 



The limits of integration are now from x = 0, the origin, to x = 2 a cos 2 cr, 

 the value of x on the plane PQ. Evaluating the integrals, and substituting 

 these limits, we obtain 



__ 2 a-rr | (2 a cos 2 a) 8 - IT sec 2 a j- (2 a cos 2 or) 4 

 ~~ 2 a?r i (2 a cos 2 a) 2 - TT sec 2 a $ (2 a cos 2 a) 8 

 = a cos 2 a, 

 giving the same result as before. 



