Solução_Calculo_Stewart_6e

F.
302 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
3. (a) See Definition 17.2.2 [ ET 16.2.2].
(b) We normally evaluate the line integral using Formula 17.2.3 [ ET 16.2.3].
(c) The mass is m = C ρ (x, y) ds,andthecenterofmassis(x, y) where x = 1 m
C xρ (x, y) ds, y = 1 m
(d) See (5) and (6) in Section 17.2 [ ET 16.2] for plane curves; we have similar definitions when C is a space curve
(see the equation preceding (10) in Section 17.2 [ ET 16.2]).
(e) For plane curves, see Equations 17.2.7 [ ET 16.2.7]. We have similar results for space curves
(see the equation preceding (10) in Section 17.2 [ ET 16.2]).
4. (a) See Definition 17.2.13 [ ET 16.2.13].
(b) If F is a force field, F · dr represents the work done by F in moving a particle along the curve C.
C
(c) F · dr = Pdx+ Qdy+ Rdz
C C
5. See Theorem 17.3.2 [ ET 16.3.2].
C
yρ (x, y) ds.
6. (a) F · dr is independent of path if the line integral has the same value for any two curves that have the same initial and
C
terminal points.
(b) See Theorem 17.3.4 [ ET 16.3.4].
7. See the statement of Green’s Theorem on page 1091 [ ET 1055].
8. See Equations 17.4.5 [ ET 16.4.5].
∂R
9. (a) curl F =
∂y − ∂Q ∂P
i +
∂z ∂z − ∂R ∂Q
j +
∂x ∂x − ∂P
k = ∇ × F
∂y
(b) div F = ∂P
∂x + ∂Q
∂y + ∂R
∂z = ∇ · F
(c) For curl F, see the discussion accompanying Figure 1 on page 1100 [ ET 1064] as well as Figure 6 and the accompanying
discussion on page 1132 [ ET 1096]. For div F, see the discussion following Example 5 on page 1102 [ ET 1066] as well
as the discussion preceding (8) on page 1139 [ ET 1103].
10. See Theorem 17.3.6 [ ET 16.3.6]; see Theorem 17.5.4 [ ET 16.5.4].
11. (a) See (1) and (2) and the accompanying discussion in Section 17.6 [ ET 16.6] ; See Figure 4 and the accompanying
discussion on page 1107 [ ET 1071] .
(b) See Definition 17.6.6 [ ET 16.6.6 ].
(c) See Equation 17.6.9 [ ET 16.6.9].
12. (a)See(1)inSection17.7[ET16.7].
(b) We normally evaluate the surface integral using Formula 17.7.2 [ ET 16.7.2].
(c) See Formula 17.7.4 [ ET 16.7.4].
TX.10
(d) The mass is m = ρ(x, y, z) dS and the center of mass is (x, y, z) where x = 1 S m
y = 1 yρ(x, y, z) dS, z =
1
m S m
zρ(x, y, z) dS.
S
S
xρ(x, y, z) dS,