436-439] Stokes' Theorem 377 



439. To prove the theorem, let us select any two points A, B on the 

 contour, and let us introduce a quantity / defined by 



r f" 8 / Y ^ x v^y 17 dz 

 } A \ ds ds ds 



the path from A to B being the same as that followed in the integral of 

 equation (363). Let us also introduce a quantity J equal to the same 

 integral taken from A to B, but along the opposite edge of the shell. Then 

 the whole integral on the left of equation (363) is equal to / J. 



FIG. 111. 



It will be possible to connect A and B by a series of non-intersecting 

 lines drawn in the shell in such a way as to divide the whole shell into 

 narrow strips. Let us denote these lines by the letters a, b, ... n, the lines 

 being taken in order across the shell, starting with the line nearest to that 

 along which we integrate in calculating /. Let us denote the value of 



ds ds 

 taken along the line a by 7 a . 



Then the left-hand member of equation (363) 



Let us consider the value of any term of this series, say I a /&. 



Let us take each point on the line a and cause it to undergo a slight 

 displacement, so that the coordinates of any point so, y, z are changed to 

 x + Sx, y + 8y, z 4- Sz. If &c, By, Bz are continuous functions of x, y, z the 

 result will be to displace the line a into some adjacent position, and by a 

 suitable choice of the values of &x, By, Sz this displaced position of line a can 

 be made to coincide with line b. If this is done, it is clear that the value of 



