175-177] Greeris Theorem 155 



where u p , U Q , U R , ... are the values of u at P, Q, R, .... Also, since the pro- 

 jection of each of the areas dS P , dS Q , ... on the plane of yz is dydz, we have 



dydz = lpdS P = l Q dS Q = l R dS R = . . . , 



where lp,l Q ,l R , ... are the values of I at P, Q, R, .... The signs in front of 

 IP, Q> IR> are alternately positive and negative, because, as we proceed 

 along PQR..., the normal drawn into the space between the surfaces makes 

 angles which are alternately acute and obtuse with the positive axis of x. 



FIG. 53. 

 Thus 



dydz \ ^- dx = dydz ( u p + U Q U R 



J ooc 



l Q u Q dS Q l R u R dS R ............ (95), 



and on adding the similar equations obtained for all the prisms we obtain 



..................... (96), 



the terms on the right-hand sides of equations of the type (95) combining so 

 as exactly to give the term on the right-hand side of (96). 



We can treat the functions v and w similarly, and so obtain altogether 



dv 



~ i ; - ) dxdydz = 2 I \(lu + mv + nw) dS, 

 oy dzj JJ^ 



proving the theorem. 



177. If u, v, w are the three components of any vector F, then the 

 expression 



du dv dw 



dx dy dz 



is denoted, for reasons which will become clear later, by div F. If N is the 

 component of the vector in the direction of the normal (I, m, n) to dS, then 



N =lu + mv + nw. 



