Solução_Calculo_Stewart_6e

F.
TX.10
(b) xy dx + C x2 y 3 dy =
∂
D ∂x (x2 y 3 ) − ∂ (xy) ∂y
dA = 1
2x
0 0 (2xy3 − x) dy dx
= 1
1
0 2 xy4 − xy y=2x
dx = 1
y=0 0 (8x5 − 2x 2 ) dx = 4 − 2 = 2 3 3 3
SECTION 17.4 GREEN’S THEOREM ET SECTION 16.4 ¤ 279
5. The region D enclosed by C is given by {(x, y) | 0 ≤ x ≤ 2,x≤ y ≤ 2x},so
C xy2 dx +2x 2 ydy =
∂
D ∂x (2x2 y) − ∂
∂y (xy2 ) dA
= 2
0
= 2
0
2x
x
(4xy − 2xy) dy dx
xy
2 y=2x
dx
y=x
= 2
0 3x3 dx = 3 4 x4 2
0 =12
7.
C
y + e √
x
dx +(2x +cosy 2 ) dy = D
∂
(2x ∂x +cosy2 ) − ∂
∂y
y + e √
x
dA
= 1 √ y
(2 − 1) dx dy = 1
0 y 2 0 (y1/2 − y 2 ) dy = 1 3
9.
C y3 dx − x 3 dy =
∂
D ∂x (−x3 ) − ∂
∂y (y3 ) dA = D (−3x2 − 3y 2 ) dA = 2π
2
0 0 (−3r2 ) rdrdθ
= −3 2π
0
dθ 2
0 r3 dr = −3(2π)(4) = −24π
11. F(x, y) = √ x + y 3 ,x 2 + √ y and the region D enclosed by C is given by {(x, y) | 0 ≤ x ≤ π, 0 ≤ y ≤ sin x}.
C is traversed clockwise, so −C gives the positive orientation.
F · d r = − √
C −C x + y
3
dx + x 2 + √ y dy = −
∂
x 2 + √
y − ∂
D ∂x
∂y x + y
3
dA
= − π
sin x
(2x − 3y 2 ) dy dx = − π
0 0 0 2xy − y
3 y=sin x
dx
= − π
0 (2x sin x − sin3 x) dx = − π
0 (2x sin x − (1 − cos2 x)sinx) dx
= − 2sinx − 2x cos x +cosx − 1 3 cos3 x π
= −
2π − 2+ 2 3 =
4
− 2π 3
y=0
0
[integrate by parts in the first term]
13. F(x, y) = e x + x 2 y, e y − xy 2 and the region D enclosed by C is the disk x 2 + y 2 ≤ 25.
C is traversed clockwise, so −C gives the positive orientation.
F · d r = − C −C (ex + x 2 y) dx +(e y − xy 2 ) dy = −
∂
D ∂x (ey − xy 2 ) − ∂
∂y (ex + x 2 y) dA
= − D (−y2 − x 2 ) dA = D (x2 + y 2 ) dA = 2π
5
0 0 (r2 ) rdrdθ
= 2π
dθ 5
0 0 r3 dr =2π 1
r4 5
= 625 π
4 0 2