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870 Chapter 27 Line integrals and multiple integrals
If the coordinates of the end points of the curveC are known, say (x 1
,y 1
) and (x 2
,y 2
),
we often write ∫ (x 2
,y 2
)
(x 1
,y 1
)
to show this, but care must then be taken to define the intended
route from AtoB.
Wenowexplainhowtointegrateafunctionalongacurve.Consideravectorfield,F,
through which runs a curve,C, as shown inFigure 27.3.
F(x,y)
F
F n
M
u
F t
N
C
MN 5 ds
Figure27.3∫
Theintegral F · dsis equal to the work
C
done by Fasthe particle movesalongC.
Suppose we restrict ourselves to two-dimensional situations. In the general case the
vector field will vary asxandyvary, that is F = F(x,y). Consider the small element
ofC joining points M and N. Let θ be the angle between the tangent to the curve at M
andthedirectionofthefieldthere.WeshalldenotethevectorjoiningMandN(i.e. −→ MN)
by δs. Consider the quantity
F·δs
where · represents the scalar product. When F represents a gravitational force field,
F·δsrepresentsthesmallamountofworkdonebythefieldinmovingaparticleofunit
mass from M to N. The appropriate integral along the whole curve represents the total
work done. From the definition of the scalar product wenote that
F·δs = |F‖δs|cosθ
Writingthe modulus ofFas simplyF, and the modulus of δs, as δs, we have
F·δs=Fδscosθ
= (Fcosθ)δs
=F t
δs
whereF t
is the component of F tangential toC. This result is of the same form as the
expressions for work done obtained previously. We are therefore interested in integrals
of the form
∫
F(x,y)·ds
C