James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 16.2 Line Integrals 1079
vector representation of the line segment that starts at r 0 and ends at r 1 is given by
8 rstd − s1 2 tdr 0 1 t r 1 0 < t < 1
(See Equation 12.5.4.)
C¡
y
(0, 2)
C
0 4
x
ExamplE 4 Evaluate y C
y 2 dx 1 x dy, where (a) C − C 1 is the line segment from
s25, 23d to s0, 2d and (b) C − C 2 is the arc of the parabola x − 4 2 y 2 from s25, 23d
to s0, 2d. (See Figure 7.)
SOLUTION
(a) A parametric representation for the line segment is
x=4-¥
x − 5t 2 5 y − 5t 2 3 0 < t < 1
(_5, _3)
FIGURE 7
(Use Equation 8 with r 0 − k25, 23l and r 1 − k0, 2l.) Then dx − 5 dt, dy − 5 dt, and
Formulas 7 give
y y 2 dx 1 x dy −
C1
y 1
s5t 2 3d 2 s5 dtd 1 s5t 2 5ds5 dtd
0
− 5 y 1
s25t 2 2 25t 1 4d dt
0
− 5F 25t 3
3
2 25t 2
2
1
1 − 2 5 4tG0 6
(b) Since the parabola is given as a function of y, let’s take y as the parameter and
write C 2 as
x − 4 2 y 2 y − y 23 < y < 2
Then dx − 22y dy and by Formulas 7 we have
y
C2
y 2 dx 1 x dy − y 2 23
y 2 s22yd dy 1 s4 2 y 2 d dy
− y 2 s22y 3 2 y 2 1 4d dy
23
−F2 y 4
2 2 y 3
3 1 4y − 40 5 6 G23
2
■
Notice that we got different answers in parts (a) and (b) of Example 4 even though the
two curves had the same endpoints. Thus, in general, the value of a line integral depends
not just on the endpoints of the curve but also on the path. (But see Section 16.3 for conditions
under which the integral is independent of the path.)
Notice also that the answers in Example 4 depend on the direction, or orientation, of
the curve. If 2C 1 denotes the line segment from s0, 2d to s25, 23d, you can verify, using
the parametrization
x − 25t y − 2 2 5t 0 < t < 1
that y 2C1
y 2 dx 1 x dy − 5 6
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