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

27.3 Evaluation of line integrals in two dimensions 871
ds
dxi
Figure27.4
Thevector dscan be
written asdxi +dyj.
dyj
Since F is a vector function of x and y it will have Cartesian components P(x,y) and
Q(x,y) and so wecan writeFinthe form
F(x,y) =P(x,y)i +Q(x,y)j
Similarly,referringtoFigure27.4, wecanwritethevectordsasds = dxi +dyj.Hence
∫ ∫
F(x,y)· ds = (P(x,y)i +Q(x,y)j)· (dxi +dyj)
C
Taking the scalar product gives
∫
P(x,y)dx +Q(x,y)dy
C
and thisisthe lineintegral inCartesian form.
C
IfF(x,y) =P(x,y)i +Q(x,y)jthen
∫ ∫
F(x,y)· ds = (P(x,y)i +Q(x,y)j)· (dxi +dyj)
C
C
∫
= P(x,y)dx +Q(x,y)dy
C
In order to proceed further we now need to explore how such integrals are evaluated
inpractice. This isthe topic of the next section.
27.3 EVALUATIONOFLINEINTEGRALSINTWODIMENSIONS
Example27.1 Evaluate
∫
F·ds
C
whereF = 5y 2 i +2xyj andC isthe straightline joiningthe originand the point (1,1).
Solution We compare the integrand with the standard form and recognize that in this case
P(x,y) = 5y 2 andQ(x,y) = 2xy. The integralbecomes
∫
5y 2 dx +2xydy
C
The curve of interest, in this case a straight line, is shown in Figure 27.5. Along this
curve it is clear that y = x at all points. We use this fact to simplify the integral by
writing everything in terms of x. We could equally well choose to write everything in
terms of y. If y = x then dy = 1, that is dy = dx. As we move along the curveC, x
dx
ranges from0to1,and the integral reduces to
∫
∫ ∫ x=1
[ 7x
5x 2 dx +2x 2 dx = 7x 2 dx = 7x 2 3
dx = = 7 3 3
C
C
x=0
] 1
0