Mathematical Methods for Physics and Engineering - Matematica.NET

LINE, SURFACE AND VOLUME INTEGRALS11.8.3 Physical applications of the divergence theoremThe divergence theorem is useful in deriving many of the most important partialdifferential equations in physics (see chapter 20). The basic idea is to use thedivergence theorem to convert an integral form, often derived from observation,into an equivalent differential form (used in theoretical statements).◮For a compressible fluid with time-varying position-dependent density ρ(r,t) and velocityfield v(r,t), in which fluid is neither being created nor destroyed, show that∂ρ+ ∇ · (ρv) =0.∂tFor an arbitrary volume V in the fluid, the conservation of mass tells us that the rate ofincrease or decrease of the mass M of fluid in the volume must equal the net rate at whichfluid is entering or leaving the volume, i.e.∮dM= − ρv · dS,dt Swhere S is the surface bounding V . But the mass of fluid in V is simply M = ∫ ρdV,soVwe have∫ ∮dρdV + ρv · dS =0.dt VSTaking the derivative inside the first integral on the RHS and using the divergence theoremto rewrite the second integral, we obtain∫∂ρV ∂t∫VdV + ∇ · (ρv) dV =∫V[ ∂ρ∂t + ∇ · (ρv) ]dV =0.Since the volume V is arbitrary, the integrand (which is assumed continuous) must beidentically zero, so we obtain∂ρ+ ∇ · (ρv) =0.∂tThis is known as the continuity equation. It can also be applied to other systems, forexample those in which ρ is the density of electric charge or the heat content, etc. For theflow of an incompressible fluid, ρ = constant and the continuity equation becomes simply∇ · v =0.◭In the previous example, we assumed that there were no sources or sinks inthe volume V , i.e. that there was no part of V in which fluid was being createdor destroyed. We now consider the case where a finite number of point sourcesand/or sinks are present in an incompressible fluid. Let us first consider thesimple case where a single source is located at the origin, out of which a quantityof fluid flows radially at a rate Q (m 3 s −1 ). The velocity field is given byv =Qr4πr 3 = Qˆr4πr 2 .Now, for a sphere S 1 of radius r centred on the source, the flux across S 1 is∮v · dS = |v|4πr 2 = Q.S 1404