Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 17 REVIEW ET CHAPTER 16 ¤ 303
13. (a) See Figures 6 and 7 and the accompanying discussion in Section 17.7 [ ET 16.7]. A Möbius strip is a nonorientable
surface; see Figures 4 and 5 and the accompanying discussion on page 1121 [ ET 1085].
(b) See Definition 17.7.8 [ ET 16.7.8].
(c) See Formula 17.7.9 [ ET 16.7.9].
(d) See Formula 17.7.10 [ ET 16.7.10].
14. See the statement of Stokes’ Theorem on page 1129 [ ET 1093.].
15. See the statement of the Divergence Theorem on page 1135 [ ET 1099].
16. In each theorem, we have an integral of a “derivative” over a region on the left side, while the right side involves the values of
the original function only on the boundary of the region.
1. False; div F is a scalar field.
3. True, by Theorem 17.5.3 [ ET 16.5.3] and the fact that div 0 =0.
5. False. See Exercise 17.3.33 [ ET 16.3.33]. (But the assertion is true if D is simply-connected; see Theorem 17.3.6
[ ET 16.3.6].)
7. True. Apply the Divergence Theorem and use the fact that div F =0.
1. (a) Vectors starting on C point in roughly the direction opposite to C, so the tangential component F · T is negative.
Thus F · dr = F · T ds is negative.
C C
(b) The vectors that end near P are shorter than the vectors that start near P ,sothenetflow is outward near P and
div F (P ) is positive.
3.
C yz cos xds = π
(3 cos t)(3sint)cost (1)
0 2 +(−3sint) 2 +(3cost) 2 dt = π
(9 0 cos2 t sin t) √ 10 dt
=9 √ 10 − 1 3 cos3 t π
= −3 √ 10 (−2) = 6 √ 10
0
5.
C y3 dx + x 2 dy = 1
−1 y 3 (−2y)+(1− y 2 ) 2 dy = 1
−1 (−y4 − 2y 2 +1)dy
= − 1 5 y5 − 2 3 y3 + y 1
−1 = − 1 5 − 2 3 +1− 1 5 − 2 3 +1= 4 15
7. C: x =1+2t ⇒ dx =2dt, y =4t ⇒ dy =4dt, z = −1+3t ⇒ dz =3dt, 0 ≤ t ≤ 1.
C xy dx + y2 dy + yz dz = 1
[(1 + 2t)(4t)(2) + 0 (4t)2 (4) + (4t)(−1+3t)(3)] dt
9. F(r(t)) = e −t i + t 2 (−t) j +(t 2 + t 3 ) k, r 0 (t) =2t i +3t 2 j − k and
= 1
0 (116t2 − 4t) dt = 116
3 t3 − 2t 2 1
= 116
110
− 2=
0 3 3
C F · dr = 1
0 (2te−t − 3t 5 − (t 2 + t 3 )) dt = −2te −t − 2e −t − 1 2 t6 − 1 3 t3 − 1 4 t4 1
0 = 11
12 − 4 e .