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082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

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27.7 Some simple volume and surface integrals 891

In Figure 27.19(b) we have placed an imaginary surface,S, within the flow and

it is reasonable to ask ‘what is the volume of fluid which crosses this surface in

any given time?’ Suppose we select a very small portion of the surface having area

δS. Note that δS is a scalar. We can also define an associated vector δS, which has

magnitude δS and whose direction is normal to this portion of the surface. The

component of fluid velocity in the direction of δS is given by the scalar product

v· ̂δS where ̂δS is a unit vector in the direction of δS. The volume of fluid crossing

this portion each second is v·δS. This is also known as the flux of v. If we

wishtocalculatethetotalfluxthroughthesurfaceSwemustsumallsuchcontributions

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

this as

volume flowper second =total flux = v·dS

This is an example of a surface integral, so called because we must integrate over

a surface. It will usually take the form of a double integral such as those discussed

inSection27.6.1.Technically,therearetwointegralsigns,butforbrevitythesehave

been replaced by the single ∫ where it is to be understood that the integral is to be

S

performed over the surfaceS. In any specific problem care must be taken to ensure

that the entire surface is included when the integration is performed. The double

integralsevaluatedinSection27.6arespecialcasesofsurfaceintegrals,inwhichthe

surface isaplane region (thex--y plane).

Forexample,supposethevelocityfieldisgivenbyv = (x+2y)k.Thisrepresents

a flow in the z direction whose magnitude varies with x and y. Suppose S is the

plane surface defined by 1 x 4, 0 y 2,z = 0. This surface is shown in

Figure 27.20. Note that the normal to this surface is parallel to thezaxis and so we

can writedS = dxdyk. Then

v·dS=(x+2y)k·dxdyk

=(x+2y)dxdy

S

y

S

y= 2

dS = dxdyk

v= (x+2y)k

x

y= 0

0

x = 1

z

x = 4

Figure27.20

The vectorfieldvrepresents

fluid flowing acrossthe

surfaceS.Theintegral over

the surface ofv ·dSgives the

volume flow persecond

acrossS.

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