SPHERICAL CAP AXD BELT 



131 



may be regarded as a narrow strip of length 2 IT a sin 6 and of 

 width a d6. Its area is accordingly 2 TTO? sin dO. 



When dO is made very small, the arc BA may be regarded as a 



7T 



straight line of length a dd, making an angle 6 with OE. 



Thus 



the length of la, the projection of BA on OE, is add cos - 6 

 or a sin 6 d6. The area of the ring BA is now seen to be 





= 2 Tra 2 sin 

 2 Tra 5a. 



dd 



Thus the mass of the ring is the same as the mass of the ele- 

 ment ba of a rod OE, if this rod is of uniform density such that 

 its mass per unit length is that of an area 2 Tra of the shell. The 

 center of gravity of the ring we have been considering clearly lies 

 on the axis OE, so that in finding the center of gravity of the 

 spherical cap this ring may obviously be replaced by the element 

 la of this rod. 



In the same way each small ring may be replaced by the cor- 

 responding element of the rod. Thus the whole cap may be 

 replaced by the length rE of the rod (fig. 74) which is inter- 

 cepted between the boundary-plane PQ and the sphere. Since 

 the rod is uniform, the center of gravity of the portion rE of 

 the rod is at its middle point. This point is therefore the center 

 of gravity of the spherical cap. 



102. Center of gravity of a belt cut from a 

 spherical shell by two parallel planes. In the 

 same way we can find the center of gravity of 

 the belt cut off from a uniform spherical shell 

 by two parallel planes. In fig. 76 let PQ, P'Q' 

 be the two planes. Then we can divide the 

 belt into narrow rings by planes parallel to 

 PQ. Each ring, as before, may be replaced by 

 the corresponding element of a uniform rod 

 along the axis OE, so that the whole belt may 

 be replaced by -the portion rr' of this rod, the 

 portion intercepted between the two planes PQ, 

 P r Q r . The center of gravity is now seen to be the middle point of rr f . 



FIG. 76 



