Mathematical Methods for Physics and Engineering - Matematica.NET

LINE, SURFACE AND VOLUME INTEGRALSSubstituting this into (11.26) and taking c out of both integrals because it is constant, wefind∫∮c · dS ×∇φ = c · φdr.SCSince c is an arbitrary constant vector we therefore obtain the stated result (11.24). ◭Equation (11.25) may be proved in a similar way, by letting a = b × c in Stokes’theorem, where c is again a constant vector. We also note that by setting b = rin (11.25) we find∫∮(dS ×∇) × r = dr × r.SCExpanding out the integrand on the LHS gives(dS ×∇) × r = dS − dS(∇ · r) =dS − 3 dS = −2 dS.Therefore, as we found in subsection 11.5.2, the vector area of an open surface Sis given by∫S = dS = 1 ∮r × dr.S 2 C11.9.2 Physical applications of Stokes’ theoremLike the divergence theorem, Stokes’ theorem is useful in converting integralequations into differential equations.◮From Ampère’s law, derive Maxwell’s equation in the case where the currents are steady,i.e. ∇×B − µ 0 J = 0.Ampère’s rule for a distributed current with current density J is∮∫B · dr = µ 0 J · dS,Cfor any circuit C bounding a surface S. Using Stokes’ theorem, the LHS can be transformedinto ∫ (∇×B) · dS; henceS∫S(∇×B − µ 0 J) · dS =0for any surface S. This can only be so if ∇×B − µ 0 J = 0, which is the required relation.Similarly, from Faraday’s law of electromagnetic induction we can derive Maxwell’sequation ∇×E = −∂B/∂t. ◭In subsection 11.8.3 we discussed the flow of an incompressible fluid in thepresence of several sources and sinks. Let us now consider vortex flow in anincompressible fluid with a velocity fieldSv = 1 ρêφ,in cylindrical polar coordinates ρ, φ, z. For this velocity field ∇×v equals zero408