Seção 16_4_S

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SEÇÃO 16.4 TEOREMA DE GREEN  3
16.4 SOLUÇÕES
 1. (a)
C x
2y dx + xy 3 dy = C 1+C 2+C 3+C 4 x
2y dx + xy 3 dy
= 10 0 dx +
1
0 y
3 dy + 01 x
2 dx + 01 0 dy
= 14 −
1
3 = −
1
12
(b) C x
2y dx + xy 3 dy = 10
1
0 y
3 − x 2 dx dy
= 10 y
3 − 13 dy =
1
4 −
1
3 = −
1
12
 2. (a)
C x d x − x
2y2 dy = C 1+C 2+C 3 x dx − x
2y2 dy
= 10 x − x
4 dx + 01 x dx +
0
1 0 dy
= 12 −
1
5 −
1
2 = −
1
5
(b) C x dx − x
2y2 dy = 10
1
x −2xy
2 − 0 dy dx
= 10
2
3 x
4 − x dx = 23 −
3
10 = −
1
5
 3. (a)
C (x + 2y) dx + (x − 2y) dy
= C 1 + C 2 (x + 2y) dx + (x − 2y) dy
= 10 x + 2x
2 + x − 2x 2 (2x ) dx
+ 01 [3x + (−x )] dx
= 10 x + 4x
2 − 4x 3 dx + 01 2x d x
= 12 +
4
3 − 1 − 1 = −
1
6
(b) C (x + 2y) dx + (x − 2y) dy =
1
0
x
x 2 (1 − 2) dy dx
= 10 x
2 − x dx = 13 −
1
2 = −
1
6
 4. (a)
C x
2 + y2 dx + 2xy dy
= C 1+C 2+C 3 x
2 + y2 dx + 2xy dy
= 20 x
2 + x 4 + 2x 3 (2x ) dx
+ 02 x
2 + 16 dx + 04 0 dy
= 83 + 32 −
8
3 − 32 = 0
(b) C x
2 + y2 dx + 2xy dy
=
D
∂
∂x
(2xy ) − ∂
∂y
x 2 + y2 dA
= D (2y − 2y) dA = D (0) dA = 0
 5. C xy dx + y
5 dy = 20
x/ 2
0 (0 − x ) dy dx
= 20 −
1
2 x
2 dx = − 43
 6. 1
− 1
1
− 1 y
5 − x 2 dy dx = 1− 1 −
2
3 dx = −
4
3
 7. D (0 − 0) dA = 0
 8. x 2+ y 2≤ 1 0 − x
2 dA = − 2pi0
1
0 r
3 cos2 θ dr dθ
= −pi 14 = −
pi
4
 9. D (2x − 2x ) dA = 0
 10. D (2x − x ) dA =
pi
0
sen x
0 x dy dx =
pi
0 x sen x dx
= [−x cos x + senx ]pi0 = pi
 11. 
D
∂
∂x
(3x + sen y) − ∂
∂y
y2 − tg− 1 x dA
= 2− 2
4
x 2 (3 − 2y) dy dx
= 2− 2 −4 − 3x
2 + x 4 dx = − 965
 12. A região D delimitada por C é dada por 
 (x, y) | −2 ≤ x ≤ 2,− 4 − x 2 ≤ y ≤ 4 − x 2 ou, em 
 coordenadas polares, {(r, θ) | 0 ≤ θ ≤ pi, 0 ≤ r ≤ 2}. Logo,
 
C
xy dx + 2x 2 dy =
D
∂
∂x
2x 2 − ∂
∂y
(xy ) dA
= D (4x − x ) dA =
pi
0
2
0 (3r cos θ) r dr dθ
= 3 pi0 cos θ dθ
2
0 r
2 dr = 3 [sen θ]pi0
1
3 r
3 2
0
= 3 (0) 83 = 0
Revisão técnica: Ricardo Miranda Martins – IMECC – Unicamp
4  SEÇÃO 16.4 TEOREMA DE GREEN
 13. C x
3 − y3 dx + x 3 + y3 dy
=
1 ≤ x 2 + y 2 ≤ 9
∂
∂x
x 3 + y3 − ∂
∂y
x 3 − y3 dA
= 1 ≤ x 2 + y 2 ≤ 9 3x
2 + 3y2 dA
= 3 pi− pi
3
1 r
3 dr dθ = 6pi 814 −
1
4 = 120pi
 14. A região D delimitada por C é dada, em coordenadas polares,
 
por (r, θ)| 0 ≤ θ ≤ pi4 , 0 ≤ r ≤ 2 . Assim,
C F · dr = C y
2 − x 2y dx + xy 2 dy
= D y
2 − 2y + x 2 dA
= pi/ 40
2
0 r
2 − 2r sen θ r dr dθ
= pi/ 40 4 −
16
3 sen θ dθ
= 4θ + 163 cos θ
pi/ 4
0 = pi +
8
3 2 − 2
 15. C F · dr = C y
6 dx + xy 5 dy
=
D
∂
∂x
xy 5 − ∂
∂y
y6 dA
= D −5y
5 dA = 0
 
 porque -5y5 é uma função ímpar de y e D é simétrica em 
relação ao eixo y.
 16. C F · dr = C x
3y dx + x 4 dy
= 0≤ x 4+ y 4≤ 1 4x
3 − x 3 dA
= 1− 1
4 1− y 4
− 4 1− y 4
3x 3 dx dy
= 1− 1
3
4 x
4
4 1− y 4
− 4 1− y 4
dx = 0
 17. A = C x d y =
2pi
0 cos
3 t 3 sen2 t cos t dt
= 3 2pi0 cos
4 t sen2 t dt
= 3 − 16 sen t cos
5 t
+ 16
1
4 sen t cos
3 t + 38 (cos t sen t)+
3
8 t
2pi
0
= 3 16
6
8 pi =
3
8 pi
Ou:
3 2pi0 cos
4 t sen2 t dt
= 3 2pi0
1
8
1
2 (1 − cos 4t)+ sen
2 2t cos 2t dt = 38 pi
 18. A = C x dy =
2pi
0 (cos t) 3 sen
2 t cos t dt
= 3 2pi0
1
8 (1 − cos 4t) dt =
3
4 pi