Solução_Calculo_Stewart_6e

F.
SECTION 17.3 THE TX.10 FUNDAMENTAL THEOREM FOR LINE INTEGRALS ET SECTION 16.3 ¤ 277
(b) The initial point of C is r(0) = (0, 1) and the terminal point is r(1) = (2, 1), so
F · dr = f(2, 1) − f(0, 1) = 2 − 0=2.
C
15. (a) f x(x, y, z) =yz implies f(x, y, z) =xyz + g(y, z) and so f y(x, y, z) =xz + g y(y, z). Butf y(x, y, z) =xz so
g y(y, z) =0 ⇒ g(y, z) =h(z). Thusf(x, y, z) =xyz + h(z) and f z(x, y, z) =xy + h 0 (z). But
f z(x, y, z) =xy +2z,soh 0 (z) =2z ⇒ h(z) =z 2 + K. Hence f(x, y, z) =xyz + z 2 (taking K =0).
(b) F · dr = f(4, 6, 3) − f(1, 0, −2) = 81 − 4=77.
C
17. (a) f x(x, y, z) =y 2 cos z implies f(x, y, z) =xy 2 cos z + g(y, z) and so f y(x, y, z) =2xy cos z + g y(y, z). But
f y(x, y, z) =2xy cos z so g y(y, z) =0 ⇒ g(y, z) =h(z). Thus f(x, y, z) =xy 2 cos z + h(z) and
f z (x, y, z) =−xy 2 sin z + h 0 (z). Butf z (x, y, z) =−xy 2 sin z, soh 0 (z) =0 ⇒ h(z) =K. Hence
f(x, y, z) =xy 2 cos z (taking K =0).
(b) r(0) = h0, 0, 0i, r(π) = π 2 , 0,π so F · dr = C f(π2 , 0,π) − f(0, 0, 0) = 0 − 0=0.
19. Here F(x, y) =tany i + x sec 2 y j. Thenf(x, y) =x tan y is a potential function for F,thatis,∇f = F so
F is conservative and thus its line integral is independent of path. Hence
tan ydx+ x C sec2 ydy= F · d r =f
2, π C 4 − f(1, 0) = 2 tan
π
− tan 0 = 2.
4
21. F(x, y) =2y 3/2 i +3x y j, W = C F · d r. Since∂(2y3/2 )/∂y =3 √ y = ∂(3x y )/∂x, there exists a function f such
that ∇f = F. Infact,f x(x, y) =2y 3/2 ⇒ f(x, y) =2xy 3/2 + g(y) ⇒ f y(x, y) =3xy 1/2 + g 0 (y). But
f y (x, y) =3x √ y so g 0 (y) =0or g(y) =K. We can take K =0 ⇒ f(x, y) =2xy 3/2 .Thus
W = F · d r = f(2, 4) − f(1, 1) = 2(2)(8) − 2(1) = 30.
C
23. We know that if the vector field (call it F) is conservative, then around any closed path C, F · dr =0.ButtakeC to be a
C
circle centered at the origin, oriented counterclockwise. All of the field vectors that start on C are roughly in the direction of
motion along C, so the integral around C will be positive. Therefore the field is not conservative.
25. From the graph, it appears that F is conservative, since around all closed
paths, the number and size of the field vectors pointing in directions similar
to that of the path seem to be roughly the same as the number and size of the
vectors pointing in the opposite direction. To check, we calculate
∂
∂
(sin y) =cosy = (1 + x cos y). ThusF is conservative, by
∂y ∂x
Theorem 6.
27. Since F is conservative, there exists a function f such that F = ∇f,thatis,P = f x, Q = f y ,andR = f z.SinceP ,
Q and R have continuous first order partial derivatives, Clairaut’s Theorem says that ∂P/∂y = f xy = f yx = ∂Q/∂x,
∂P/∂z = f xz = f zx = ∂R/∂x,and∂Q/∂z = f yz = f zy = ∂R/∂y.