124 CENTER OF GRAVITY 



95. To take a definite instance, let us suppose that the density 

 increases uniformly from one end to the other. Let the density at 

 A be 0, and that at B be k. If the rod is of length a, the density 



at a distance x from A will be k I - \ Thus we must put 



w 



in formula (28), and so obtain 



x 

 kl- \xdx 



(H- 



J \ 



dx 



where the integration is from x = to x = a. Dividing numerator 



k 

 and denominator by -> we obtain 



. jf 



I 



xdx 



2 



showing that the center of gravity is two thirds of the way along 

 the rod. 



96. We can use this result to obtain the center of gravity of a 

 triangle. As in 92, we divide the triangle into parallel strips, 

 and replace each strip by a particle at its middle point. The mass 

 of each particle must be that of the strip which it replaces, and 

 this is jointly proportional to the width and the length of the 

 strip. If x is the distance of any particle from A measured along 

 the median AD, the width of a strip is proportional to dx, the 

 length intercepted on the median, while the length of the strip is 

 proportional to x, the distance from a. Thus p dx must be simply 





