19 



and 



q(A) =-L [3(A . n)n - A] = ^(^ ' '^)^ --^ 



R 



R' ^' 



where n has the same meaning as in Equation [22a]. Then 





R)ndo 



and by Gauss' theorem, remembering that n is directed outward 



I R(q-n)tia =4 r V{k-R)dn 

 R^] 



3 



[35] 



[36] 



[37] 



Similarly, / $nc?a depends only upon A, so 



I (Dn<^a=J_ j (A • R)nc?a = iir A 

 •^5, R] ''s. 3 



by Equation [34]. The remaining integral J q • n da depends only upon o° and is simply the 



— i 



total flow from a source of strength a°, so 



, q • " da= iff^o 



[38] 



Si 



Then, since — - = - r ■ x 

 dt ' 



F^(i) ^-ijTf) 



° ' ' dt dt 



[39] 



The extension to continuous distributions is evident: 



F, = -4;7p [*f-a°(r. X 6.) + r.^+lk 

 2 J L ' dt dt_ 



dr 



[39 T 



