439-441] Stokes' Theorem 379 



ndS, while the coordinates of three of its angular points will be x, y; x + dx, 

 y + dy\ and x + &x, y -f By. Using the usual formula for the area, we obtain 



ndS = (By dx Sx dy), 

 and using this relation in expression (364), we obtain 



the integral denoting summation over all those elements of area of the shell 

 which lie between lines a and b. By summation of three equations of the 

 type of (365), we obtain 



JL -j \AJ\J W I f^ i VVO 



4 ds } A ds 



'?*-.<M\ m dS ( - 

 ,dz dx) \dx dy. 



where the integration has the same meaning as before. If we add a system 

 of equations of this type, one for each strip, the left-hand, as already seen, 

 becomes I J, which is equal to the left-hand member of equation (363), 

 while the right-hand member of the new equation is also the right-hand 

 member of equation (363). This proves the theorem. 



440. Stokes' Theorem enables us to transform any line integral taken 

 round a closed circuit into a surface integral taken over any area by which 

 the circuit can be filled up. The converse operation of changing a surface 

 integral into a line integral may or may not be possible. 



441. THEOREM. It will be possible to transform the surface integral 



\(lu + mv + nw) dS , (366) 



into a line integral taken round the contour of the area S if, and only if, 



|? + |? + !? = () (367), 



at every point of the area S. 



It is easy to see that this condition is a necessary one. Let S' denote any 

 area having the same boundary as S, and being adjacent to it, but not 

 coinciding with it. Then if / is the line integral into which the surface 

 integral can be transformed, we must have 



(368), 



jj 

 and also 



(369). 



