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

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27.5 Conservative fields and potential functions 877

(b) Tofind thelineintegralweshallchoosethestraightlinepath,C,joining (0,0) and

(3,2). This has equationy = 2 x. The lineintegral becomes

3

∫ (3,2)

(0,0)

F·ds=

=

=

=

∫ (3,2)

(0,0)

∫ (3,2)

(0,0)

∫ (3,2)

(0,0)

∫ x=3

x=0

[ ] 7 3

=

3 x 0

= 7

(i +2j)·(dxi +dyj)

dx+2dy

dx + 4 3 dx

7

3 dx

since dy = 2 3 dxonC

(c) The value of φ at B(3, 2) is 3 + 2(2) = 7. The value of φ at A(0, 0) is 0. Clearly

the difference between these values is 7, the same as the value of the line integral

obtained inpart(b).

Theresultobtained inthe previousexample isimportant:

IfFisaconservative vectorfield suchthat F = ∇φ then, forany points Aand B,

∫ B

A

F· ds = φ(B)−φ(A)

27.5.2 Thelineintegralaroundaclosedloop

AsecondimportantpropertyofconservativefieldsariseswhenthecurveCofintegration

forms a closed loop. Suppose we evaluate a line integral from A to B firstly along the

curveC 1

and secondly alongC 2

asshown inFigure 27.7.

If the field,F, isconservative bothintegrals will yield the same answer, thatis

∫ ∫

F·ds= F·ds

C 1

C 2

But

∫ ∫

F·dsfromAtoB =−

F·dsfromBtoA

Consequently,thelineintegralaroundtheclosedcurveAtoBandbacktoAmustequal

zero.When the pathofintegration isaclosedcurve weuse the symbol

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