PYEAMID 133 



parallel planes meeting AG in g, g' respectively. Let Ag = x and 

 Ag f = x + dx, so that the lamina intercepts a length dx on ^4 G. 



Let be the angle between AG and the perpendicular from A 

 on to the base of the lamina. Then the thickness of the lamina 



= gg 1 cos 6 = dx cos 9. 



If S is the area of the base of the pyramid, the area of the 

 lamina under discussion is 



for the areas of the different laminas are proportional to the squares 

 of their linear dimensions. Thus the volume of the lamina we are 

 considering 2 



= - - S dx cos 6. 



If this is to be replaced by a particle occupying the length dx 

 of the rod AG, the density of the rod must be 



AG 2 



Thus the rod AG must be of a density which varies as the 

 square of the distance (x) from the end (A). 



The distance of the center of gravity of this rod from A is now, 

 by the formula of 94, 



C 



I pxdx 



JA 



x = 



c 



I 



JA 







pdx 

 A 



/"* " 



I x 3 dx 

 JA 



L 



