Solução_Calculo_Stewart_6e

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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.
F. 278 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16 TX.10 29. D = {(x, y) | x>0, y>0} = the first quadrant (excluding the axes). (a) D is open because around every point in D we can put a disk that lies in D. (b) D is connected because the straight line segment joining any two points in D lies in D. (c) D is simply-connected because it’s connected and has no holes. 31. D = (x, y) | 1
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