Seção 15_2_S

Prévia do material em texto

SEÇÃO 15.2 INTEGRAIS ITERADAS  3
 1. 20 x
2y3 dy = x 2 14 y
4 2
0 = 4x
2 ,
1
0 x
2y3 dx = y3 13 x
3 1
0 =
1
3 y
3
 2. 20 2xy − 3x
2 dy = xy 2 − 3x 2y 20 = 4x − 6x
2 ,
1
0 2xy − 3x
2 dx = yx 2 − x 3 10 = y − 1
 3. 20 xe
x + y dy = xe x [ey ]20 = x e
x +2 − ex
= xe x e2 − 1
1
0 xe
x + y dx = ey 10 xe
x dx = ey [xe x − ex ]10 = e
y
 4. 
2
0
x
y2 + 1
dy = x tg− 1 y 20 = x tg
− 1 2,
1
0
x
y2 + 1
dx = 1
y2 + 1
x 2
2
1
0
=
1
2 (y2 + 1)
 5. 40
2
0 x y dx dy =
4
0 y
1
2 x
2 2
0 dy =
4
0 2 y dy
= 43 y
3 / 2
4
0
= 323
 6. 20
3
0 e
x − y dy dx = 20 −e
x − y 3
0 dx
= 20 e
x 1 − e− 3 dx = e2 − e− 1 − 1+ e− 3
 7. 1
− 1
1
0 x
3 y2 + 3xy 2 dy dx
= 1− 1
1
4 x
3 y4 + xy 3 y=1y=0 dx =
1
− 1
1
4 x
3 + x dx
= 116 x
4 + 12 x
2 1
− 1 = 0
Solução Alternativa: Aplicando o Teorema de Fubini, a
integral torna-se igual a
1
0
1
− 1 x
3 y2 + 3xy 2 dx dy
= 10
1
4 y
2 x 4 + 32 y
2 x 2 x =1x =− 1 dy =
1
0 0 dy = 0
 8. 10
2
1 x
4 − y2 dx dy = 21
1
0 x
4 − y2 dy dx
= 21 x
4y − 13 y
3 y=1
y=0 dx =
2
1 x
4 − 13 dx
= 15 x
5 − 13 x
2
1 =
88
15
 9. pi/40
3
0 sen x dy dx = 3
pi/4
0 sen x dx = 3 [− cos x ]
pi/4
0
= 3 1 − 12
 10. pi/20
pi/2
0 sen x cos y dy dx
= pi/20 sen x d x
pi/ 2
0 cos y dy (como no Exemplo 5)
= [− cos x ]pi/20 [sen y]
pi/ 2
0 = − (0 − 1) (1 − 0) = 1
 11. 30
1
0 x + y dx dy =
3
0
2
3 (x + y)
3 / 2 x =1
x =0
dy
= 23
3
0 (1 + y)
3 / 2 − y3 / 2 dy
= 23
2
5 (1 + y)
5 / 2 − 25 y
5 / 2
3
0
= 415 32 − 3
5 / 2 − 1 = 415 31 − 9 3
 12. pi/ 2
0
pi/2
0 sen (x + y) dy dx
= pi/ 20 [− cos (x + y)]
y=pi/ 2
y=0 dx
= pi/ 20 cos x − cos x +
pi
2 dx
= sen x − sen x + pi2
pi/2
0
= (1 − 0) − (0 − 1) = 2
 13. 
2
1
3
0 2y
2 − 3xy 3 dy dx = 21
2
3 y
3 − 34 xy
4 y=3
y=0 dx
= 21 18 −
243
4 x dx = 18x −
243
8 x
2 2
1 = −
585
8
 14. 32
0
− 1 xy
2 + yx − 1 dy dx
= 32
1
3 xy
3 + 12 y
2x− 1 y=0y=− 1 dx =
3
2
1
3 x −
1
2 x
− 1 dx
= 16 x
2 − 12 ln x
3
2 =
5
6 + ln
2
3
 15. pi/60
4
1 x sen y dx dy =
pi/6
0 sen y dy
4
1 x dx
= 1 − 32
15
2 =
15 (2− 3)
4
 16. 
1
0
1
1+ y
dy
2
− 1
(1 + x ) dx
= [ln (1 + y)]10 x +
1
2 x
2 2
− 1
= (ln 2) 2+ 2+ 1 − 12 =
9
2 ln 2
 17. R xye
y dA = 20
1
0 xye
y dy dx = 20 x d x
1
0 ye
y dy
= 12 x
2 2
0 [e
y (y − 1)]10 (por partes)
= 12 (4 − 0) 0+ e
0 = 2
 18. 1
0
1
0 xe
xy dy dx = 10 [e
xy ]y=1y=0 dx =
1
0 (e
x − 1) dx
= [ex − x ]10 = e − 2
 19. 
1
0
2
1
1
x + y
dx dy = 10 [ln ( x + y)]
x =2
x =1 dy
= 10 [ln ( 2+ y) − ln (1 + y)] dy
= [(2 + y) ln ( 2 + y) − (2 + y)]
− [(1+ y) ln ( 1+ y) − (1 + y)]
1
0
por partes, separadamente, para cada
termo ou pela Tabela de Integrais
= (3 ln 3) − 3 − (2 ln 2) + 2− [(2 ln 2 − 2) − (0 − 1)]
= 3 ln 3 − 4 ln 2 = ln 2716
 20. V = R (2x + 5y + 1) dA =
4
1
0
− 1 (2x + 5y + 1) dx dy
= 41 x
2 + 5xy + x x =0x =− 1 dy =
4
1 5y dy =
5
2 y
2 4
1
= 752
15.2 SOLUÇÕES Revisão técnica: Ricardo Miranda Martins – IMECC – Unicamp
4  SEÇÃO 15.2 INTEGRAIS ITERADAS
 21. V = R x
2 + y2 dA = 3− 3
2
− 2 x
2 + y2 dx dy
= 3− 3
1
3 x
3 + y2x x =2x =− 2 dy =
3
− 3
16
3 + 4y
2 dy
= 163 y +
4
3 y
3 3
− 3 = 2 (16 + 36) = 104
 22. V = 31
1
− 1 y
2 − x 2 dx dy = 2 31
1
0 y
2 − x 2 dx dy
= 2 31 y
2x − 13 x
3 x =1
x =0 dy = 2
3
1 y
2 − 13 dy
= 23 y
3 − y 31 = 16
 23. A (R ) = pi2 · 1 =
pi
2 , assim
f med=
1
A (R ) R
f (x, y) dA
=
1
pi/2
pi/ 2
0
1
0
x sen xy dy dx
= 2pi
pi/2
0 [− cos xy ]
y=1
y=0 dx
= 2pi
pi/2
0 (1 − cos x ) dx
= 2pi [x − sen x ]
pi/2
0 = 1 −
2
pi