Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.2 LINE INTEGRALS ET SECTION 16.2 ¤ 273
17. (a) Along the line x = −3, the vectors of F have positive y-components, so since the path goes upward, the integrand F · T is
always positive. Therefore C 1
F · dr = C 1
F · T ds is positive.
(b) All of the (nonzero) field vectors along the circle with radius 3 are pointed in the clockwise direction, that is, opposite the
direction to the path. So F · T is negative, and therefore C 2
F · dr = C 2
F · T ds is negative.
19. r(t) =11t 4 i + t 3 j,soF(r(t)) = (11t 4 )(t 3 ) i +3(t 3 ) 2 j =11t 7 i +3t 6 j and r 0 (t) =44t 3 i +3t 2 j.Then
C F · dr = 1
F(r(t)) · 0 r0 (t) dt = 1
0 (11t7 · 44t 3 +3t 6 · 3t 2 ) dt = 1
0 (484t10 +9t 8 ) dt = 44t 11 + t 9 1
=45. 0
21.
C F · dr = 1
0 sin t 3 , cos(−t 2 ),t 4 · 3t 2 , −2t, 1 dt
= 1
0 (3t2 sin t 3 − 2t cos t 2 + t 4 ) dt = − cos t 3 − sin t 2 + 1 t5 1
= 6 − cos 1 − sin 1
5 0 5
23. F(r(t)) = (e t ) e −t2
i +sin e −t2
j = e t−t2 i +sin e −t2 j, r 0 (t) =e t i − 2te −t2 j.Then
C
F · dr =
=
2
1
2
25. x = t 2 , y = t 3 , z = t 4 so by Formula 9,
1
F(r(t)) · r 0 (t) dt =
e 2t−t2 − 2te −t2 sin
2
1
e t−t2 e t +sin e −t2
· −2te −t2 dt
e −t2 dt ≈ 1.9633
C x sin(y + z) ds = 5
0 (t2 )sin(t 3 + t 4 ) (2t) 2 +(3t 2 ) 2 +(4t 3 ) 2 dt
= 5
0 t2 sin(t 3 + t 4 ) √ 4t 2 +9t 4 +16t 6 dt ≈ 15.0074
27. We graph F(x, y) =(x − y) i + xy j and the curve C. We see that most of the vectors starting on C point in roughly the same
direction as C, so for these portions of C the tangential component F · T is positive. Although some vectors in the third
quadrant which start on C point in roughly the opposite direction, and hence give negative tangential components, it seems
reasonable that the effect of these portions of C is outweighed by the positive tangential components. Thus, we would expect
F · dr = F · T ds to be positive.
C C
To verify, we evaluate C F · dr. ThecurveC can be represented by r(t) =2cost i +2sint j, 0 ≤ t ≤ 3π 2 ,
so F(r(t)) = (2 cos t − 2sint) i +4cost sin t j and r 0 (t) =−2sint i +2cost j. Then
C F · dr = 3π/2
0
F(r(t)) · r 0 (t) dt
= 3π/2
0
[−2sint(2 cos t − 2sint)+2cost(4 cos t sin t)] dt
=4 3π/2
(sin 2 t − sin t cos t +2sint cos 2 t) dt
0
=3π + 2 [using a CAS]
3