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892 Chapter 27 Line integrals and multiple integrals

The surfaceintegral which gives the flux ofvthroughSis

∫ y=2 ∫ x=4

y=0

x=1

(x +2y)dxdy

Note how the limits of integration are chosen so that contributions from the whole

surfaceareincluded.ThisintegralhasalreadybeenevaluatedinExample27.10and

found tobe 27. This represents the volume flow per second throughS.

Engineeringapplication27.6

Electriccurrentdensity

Electric current is the flow of electric charges. Figure 27.21 represents an electric

current flowing across a surfaceS.

If ρ(x,y,z)isthechargedensityatapoint,thatisthechargeperunitvolume,and

v(x,y,z) isthe velocity of the charges, thenthe quantityJdefined as

J=ρv

is called the current density and has units of amperes per square metre. Suppose

weselectaverysmallportionofthesurfacehavingarea δS.Notethat δS isascalar.

Supposewedefineavector, δS,whichhasmagnitude δSandwhichisnormaltothis

portion of surface. The component of current density in the direction of δS is given

by the scalar productJ· ̂δS. The current crossing this portion isJ·δS. If we wish to

calculate the total current,I, through the surfaceSwe must sum all such contributions

over the entire surface. This is found by integrating over the surface. We write

thisas

current,I = J·dS

S

This isan example of asurface integral.

surface S

J = J(x, y,z)

dS

small portion of surface dS

Figure27.21

Theelectric current through asurface is

obtained usingasurface integral.

Note that this surface integral involves integrating a current density with units of

amperes per square metre over an area with units of square metres. Therefore the

result is a quantity with units of amperes and so the units balance on both sides of

the equation.

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