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By Feynman, Leyton, Sands.

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15) Sab and for the flux out of V2 , Flux through S2 = C · n da + Sb Sab Note that in the second integral we have written n1 for the outward normal for Sab when it belongs to S1 , and n2 when it belongs to S2 , as shown in Fig. 3-4. Clearly, n1 = −n2 , so that C · n1 da = − Sab C · n2 da. 16) Sab If we now add Eqs. 15), we see that the sum of the fluxes through S1 and S2 is just the sum of two integrals which, taken together, give the flux through the original surface S = Sa + Sb . We see that the flux through the complete outer surface S can be considered as the sum of the fluxes from the two pieces into which the volume was broken.

32), we find that [Cx (1) − Cx (3)] ∆x = − ∂Cx ∆x ∆y. 34) The derivative can, to our approximation, be evaluated at (x, y). Similarly, for the other two terms in the circulation, we may write Cy (2) ∆y − Cy (4) ∆y = ∂Cy ∆x ∆y. 36) 3-9 Loop Γ Fig. 3-9. Some surface bounded by the loop Γ is chosen. The surface is divided into a number of small areas, each approximately a square. The circulation around Γ is the sum of the circulations around the little loops. y 3 Cy C 4 ∆y 2 C 1 Cx (x, y ) ∆x x Fig.

Or we could imagine that there are some wires running into a tiny resistor that is being heated by an electric current. We shall suppose that the heat is generated practically at a point, and let W represent the energy liberated per second at that point. We shall suppose that in the rest of the volume heat is conserved, and that the heat generation has been going on for a long time—so that now the temperature is no longer changing anywhere. The problem is: What does the heat vector h look like at various places in the metal?

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