Mathematical Methods for Physics and Engineering : A Comprehensive Guide

LINE, SURFACE AND VOLUME INTEGRALS

11.8.3 Physical applications of the divergence theorem

The divergence theorem is useful in deriving many of the most important partial

differential equations in physics (see chapter 20). The basic idea is to use the

divergence 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 densityρ(r,t)and velocity

fieldv(r,t), in which fluid is neither being created nor destroyed, show that

∂ρ

∂t

+∇·(ρv)=0.

For an arbitrary volumeVin the fluid, the conservation of mass tells us that the rate of
increase or decrease of the massMof fluid in the volume must equal the net rate at which
fluid is entering or leaving the volume, i.e.

dM

dt

=−

∮

S

ρv·dS,

whereSis the surface boundingV. But the mass of fluid inVis simplyM=

∫

VρdV,so
we have
d
dt

∫

V

ρdV+

∮

S

ρv·dS=0.

Taking the derivative inside the first integral on the RHS and using the divergence theorem
to rewrite the second integral, we obtain
∫

V

∂ρ

∂t

dV+

∫

V

∇·(ρv)dV=

∫

V

[

∂ρ

∂t

+∇·(ρv)

]

dV=0.

Since the volumeVis arbitrary, the integrand (which is assumed continuous) must be
identically zero, so we obtain
∂ρ
∂t

+∇·(ρv)=0.

This is known as thecontinuity equation. It can also be applied to other systems, for
example those in whichρis the density of electric charge or the heat content, etc. For the
flow 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 in

the volumeV, i.e. that there was no part ofVin which fluid was being created

or destroyed. We now consider the case where a finite number ofpointsources

and/or sinks are present in an incompressible fluid. Let us first consider the

simple case where a single source is located at the origin, out of which a quantity

of fluid flows radially at a rateQ(m^3 s−^1 ). The velocity field is given by

v=

Qr

4 πr^3

=

Qrˆ

4 πr^2

.

Now, for a sphereS 1 of radiusrcentred on the source, the flux acrossS 1 is
∮

S 1

v·dS=|v| 4 πr^2 =Q.