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

872 Chapter 27 Line integrals and multiple integrals
y
1 (1, 1)
y
1
(1, 1)
C
C 2
C 1
C
0 1 x
Figure27.5
Thepath ofintegration joins
(0,0)and(1,1)and has equation
y=x.
0 1 x
Figure27.6
ThepathC is made upoftwo distinct
parts.Integrationis performed
separately oneach part.
Let us now see what happens when we choose to evaluate the integral of Example 27.1
along a different path joining (0,0) and (1,1).
Example27.2 Evaluate the integral ∫ C 5y2 dx +2xydy along the curve,C, consisting of thexaxis betweenx
= 0 andx = 1 and the linex = 1 asshown inFigure 27.6.
Solution WeevaluatethisintegralintwopartsbecausethecurveCismadeupoftwopieces.The
firstpieceC 1
ishorizontal,andthesecondC 2
isvertical.Therequiredintegralisthesum
of the two separate ones. Along thexaxis,y = 0 and dy = 0. This means that both the
terms5y 2 dx and 2xydy arezero,and so the integral reducesto
∫ x=1
x=0
0dx=0
andsothereiszerocontributiontothefinalanswerfromthispartofthecurveC.Along
the linex = 1, the quantity dx equals zero. Hence 5y 2 dx also equals zero, and 2xydy
equals2ydy. Becauseyrangesfrom0to1the integral becomes
∫ y=1
y=0
2ydy = [y 2 ] 1 0 = 1
Note thatthis isadifferent answerfrom thatobtained inExample 27.1.
As we have seen from Examples 27.1 and 27.2 the value of a line integral can depend
upon the pathtaken. Itisthereforeessential tospecifythispathclearly.
Example27.3 Evaluatetheintegral ∫ C F(x,y)·ds,whereF(x,y) = (3x2 +y)i+(5x−y)jandCisthe
portionofthe curvey = 2x 2 between A(2,8)and B(3, 18).
∫ ∫
Solution F(x,y)·ds = (3x 2 +y)dx + (5x −y)dy
C
C
The curveC has equationy = 2x 2 for 2 x3. Along this curve we can replaceyby
2x 2 . Note also that dy = 4x and so we can replace dy with 4xdx. This will produce an
dx