Solução_Calculo_Stewart_6e

F.
304 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
11.
∂
[(1 + ∂y xy)exy ]=2xe xy + x 2 ye xy =
∂x ∂ e y + x 2 e xy and the domain of F is R 2 ,soF is conservative. Thus there
exists a function f such that F = ∇f. Thenf y (x, y) =e y + x 2 e xy implies f(x, y) =e y + xe xy + g(x) and then
f x(x, y) =xye xy + e xy + g 0 (x) =(1+xy)e xy + g 0 (x). Butf x(x, y) =(1+xy)e xy ,sog 0 (x) =0 ⇒ g(x) =K.
Thus f(x, y) =e y + xe xy + K is a potential function for F.
13. Since ∂
∂y (4x3 y 2 − 2xy 3 )=8x 3 y − 6xy 2 = ∂
∂x (2x4 y − 3x 2 y 2 +4y 3 ) and the domain of F is R 2 , F is conservative.
Furthermore f(x, y) =x 4 y 2 − x 2 y 3 + y 4 is a potential function for F. t =0corresponds to the point (0, 1) and t =1
corresponds to (1, 1),so F · dr = f(1, 1) − f(0, 1) = 1 − 1=0.
C
15. C 1 : r(t) =t i + t 2 j, −1 ≤ t ≤ 1;
C 2 : r(t) =−t i + j, −1 ≤ t ≤ 1.
Then
C xy2 dx − x 2 ydy= 1
−1 (t5 − 2t 5 ) dt + 1
−1 tdt
Using Green’s Theorem, we have
C
= − 1 6 t6 1
−1 + 1
2 t2 1
−1 =0
∂
xy 2 dx − x 2 ydy=
D ∂x (−x2 y) − ∂
1
1
∂y (xy2 ) dA = (−2xy − 2xy) dA =
D
−1
17.
C x2 ydx− xy 2 dy =
= 1
−1
x 2 + y 2 ≤ 4
−4xy dy dx
x 2
−2xy
2 y=1
dx = 1
y=x 2 −1 (2x5 − 2x) dx = 1
3 x6 − x 2 1
=0 −1
∂
∂x (−xy2 ) − ∂
∂y (x2 y) dA = (−y 2 − x 2 ) dA = − 2π 2
0 r3 dr dθ = −8π
x 2 + y 2 ≤ 4
19. If we assume there is such a vector field G,thendiv(curl G) =2+3z − 2xz. Butdiv(curl F) =0for all vector fields F.
Thus such a G cannot exist.
21. For any piecewise-smooth simple closed plane curve C bounding a region D, we can apply Green’s Theorem to
F(x, y) =f(x) i + g(y) j to get f(x) dx + g(y) dy =
∂
g(y) − ∂ f(x) dA = 0 dA =0.
C D ∂x ∂y D
23. ∇ 2 f =0means that ∂2 f
∂x 2 + ∂2 f
∂y 2
=0.NowifF = f y i − f x j and C is any closed path in D,thenapplyingGreen’s
Theorem, we get
F · dr = f C C y dx − f x dy =
∂
(−f D ∂x x) − ∂ (f ∂y y) dA = − (f D xx + f yy ) dA = − 0 dA =0
D
Therefore the line integral is independent of path, by Theorem 17.3.3 [ ET 16.3.3].
25. z = f (x, y) =x 2 +2y with 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x. Thus
A(S) = D
√
1+4x2 +4dA = 1
0
2x
0
√
5+4x2 dy dx = 1
0 2x √ 5+4x 2 dx = 1 6 (5 + 4x2 ) 3/2 1
0
= 1 6
27 − 5
√
5
.
0