LINE, SURFACE AND VOLUME INTEGRALS
Substituting this into (11.26) and takingcout of both integrals because it is constant, we
find
c·∫
SdS×∇φ=c·∮
Cφdr.Sincecis an arbitrary constant vector we therefore obtain the stated result (11.24).
Equation (11.25) may be proved in a similar way, by lettinga=b×cin Stokes’theorem, wherecis again a constant vector. We also note that by settingb=r
in (11.25) we find
∫
S(dS×∇)×r=∮Cdr×r.Expanding 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 surfaceS
is given by
S=∫SdS=1
2∮Cr×dr.11.9.2 Physical applications of Stokes’ theoremLike the divergence theorem, Stokes’ theorem is useful in converting integral
equations into differential equations.
From Amp`ere’s law, derive Maxwell’s equation in the case where the currents are steady,
i.e.∇×B−μ 0 J= 0.Ampere’s rule for a distributed current with current density` Jis
∮
CB·dr=μ 0∫
SJ·dS,for any circuitCbounding a surfaceS. Using Stokes’ theorem, the LHS can be transformed
into
∫
S(∇×B)·dS; hence ∫S(∇×B−μ 0 J)·dS=0foranysurfaceS. 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’s
equation∇×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 considervortexflow in an
incompressible fluid with a velocity field
v=1
ρeˆφ,in cylindrical polar coordinatesρ, φ, z. For this velocity field∇×vequals zero