

76, 77] NEWTONIAN POTENTIAL FUNCTION. 151 



dV 



second is continuous, we may make, as shown, -= as small as we 



ox 



please by making the sphere at P small enough. At a neighbour- 

 ing point P! draw a small sphere, and let the corresponding parts 



, W{ , dVS , dVS 



be ^ - and -= . Then we can make -r ^ as small as we please, 

 ox ox ox 



~\ ~rrr o -rr / 



and hence also the difference -= --- ^-. Hence by taking P and 



Pj near enough together, we can make the increment of - - as 



ox 



' dv. 



small as we please, or ^r is continuous. 

 dx 



77. Poisson's Equation. By Gauss's theorem ( 39 (5)), we 

 have 



when r is drawn from 0, a point within 8. Multiplying by m, a 

 mass concentrated at 0, 



(i) j)- 2 cos ( nr )dS = 



The integral 



I I x7O / / D / D \ ^7C* 



II T a& = 1 1 JL cos (.1/2^ c&o 



J J Ufl J J 



is the surface integral of the outward normal component of the 

 parameter P, or of the inward component of the force. 



The surface integral of the normal component of force in the 

 inward direction through $ is called the flux of force into S, and 

 we see that it is equal to 4?r times the element of mass within 

 S. Masses without contribute nothing to the integral. Every 



mass dm situated within S contributes to the potential at any 



point and kirdm to the flux through the surface 8. Hence the 

 whole mass, when potential is V\\\ , contributes to the 

 flux 



pdr, 



and 



