ILLUSTEATIVE EXAMPLE 



139 



ILLUSTRATIVE EXAMPLE 



A right circular cone OPQ is scooped out of a solid homogeneous sphere, the 

 vertex of the cone being on the surface of the sphere, and its axis being a diameter 

 of the sphere. It is required to find the center of gravity of the remainder. 



METHOD I. Polar coordinates. First let us use polar coordinates, taking the 

 vertex of the cone as origin, and the axis of the cone as initial line. If a is 

 the semivertical angle of the cone, the equation of the cone is 6 = a. If a 

 is the radius of the sphere, the equation of the sphere is r = 2 a cos 6. The 

 center of gravity must from symmetry lie on the axis = 0, so that 6 = 0, and 

 equation (34) becomes 



r = 



fffpr* sin 6 cos drd0d<j> 

 CCC P r*sm0drd0d<t> 



The solid is supposed to be homo- 

 geneous, so that p is a constant, and 

 may, therefore, be taken outside the 

 sign of integration in both numerator 

 and denominator. The limits of inte- 

 gration for <f> are from <f> = to = 2 TT, 

 so that this integration may be per- 

 formed in each case. Doing this, and dividing out by 2 irp, we are left with 



FIG. 81 



r = 



r 3 sin e cos drd0 



r 2 sin drd0 



We may next integrate with respect to r, the limits being r= to r = 2acos0, 

 and obtain 



C\ (2 a cos 0)* sin cos d0 



r = 



f}(2acos0) 3 sin0d0 

 I cos 5 sin d0 



T 



cos 3 0sin0d0 



The limits of integration for are obviously from = a (the cone) to 

 = - (the tangent plane to the sphere). We have 



TT * 



f 2 cos 5 sin dd = - $ [cos 6 0] 2 = fc cos 6 a, 



. " J a 



v ; 



cos 3 sin de = - \ [cos 4 0] 2 = \ cos* a. 



