Seção 15_7_S

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SEÇÃO 15.7 INTEGRAIS TRIPLAS  3
 1. 20
0
− 3
1
− 1 x
2 + yz dz dy dx
= 20
0
− 3 x
2 z + 12 yz
2 z =1
z =− 1 dy dx =
2
0
0
− 3 2x
2 dy dx
= 20 2x
2y y=0y=− 3 dx =
2
0 6x
2 dx = 2 x 3 20 = 16
1
− 1
0
− 3
2
0 x
2 + yz dx dy dz
= 1− 1
0
− 3
1
3 x
3 + xy z x =2x =0 dy dz
= 1− 1
0
− 3
8
3 + 2yz dy dz =
1
− 1
8
3 y + y
2 z y=0y=− 3 dz
= 1− 1 (8 − 9z ) dz = 8z −
9
2 z
2 1
− 1 = 16
1
− 1
2
0
0
− 3 x
2 + yz dy dx dz
= 1− 1
2
0 x
2y + 12 y
2 z y=0y=− 3 dx dz
= 1− 1
2
0 3x
2 − 92 z dx dz =
1
− 1 x
3 − 92 xz
x =2
x =0 dz
= 1− 1 (8 − 9z ) dz = 8z −
9
2 z
2 1
− 1 = 16
 2. 10
z
0
y
0 xyz dx dy dz =
1
0
z
0
1
2 y
3 z dy dz
= 10
1
8 z
5 dz = 148 z
6 1
0 =
1
48
 3. 10
2x
x
x + y
0 2xy dz dy dx
= 10
2x
x 2x
2y + 2xy 2 dy dx = 10 x
2y2 + 23 xy
3 2x
x dx
= 10
23
3 x
4 dx = 2315 x
4 1
0 =
23
15
 4. pi0
2
0
4− z2
0 z sen y dx dz dy
= pi0
2
0 z 4 − z 2 sen ydz dy
= pi0 −
1
3 4 − z
2 3 / 2
2
0
sen y dy = pi0
8
3 sen y dy
= − 83 cos y
pi
0 =
16
3
 5. 30
9− x 2
0
x
0 yz dy dz dx =
3
0
9− x 2
0
1
2 x
2 z dz dx
= 30
1
4 x
2 z 2 9− x
2
0 dx =
3
0
1
4 9x
2 − x 4 dx
= 3x 3 − 15 x
5 3
0 =
162
5
 6. 21
x
0
1− y
0 x
3y2 z dz dy dx = 21
x
0
1
2 x
3y2 z 2 z =1 − yz =0 dy dx
= 21
x
0
1
2 x
3y2 (1 − y)2 dy dx
= 21
x
0
1
2 x
3y2 − x 3y3 + 12 x
3y4 dy dx
= 21
1
6 x
3y3 − 14 x
3y4 + 110 x
3y5 y= xy=0 dx
= 21
1
6 x
6 − 14 x
7 + 110 x
8 dx
= 142 x
7 − 132 x
8 + 190 x
9 2
1
= 12842 −
256
32 +
512
90 −
1
42 +
1
32 −
1
90 =
7387
10 080
 7. 10
2z
0
z +2
0 yz dx dy dz =
1
0
2z
0 yz (z + 2) dy dz
= 10 2z
4 + 4z3 dz = 75
 8. 10
y
0
x + y
0 e
x dz dx dy = 10
y
0 (x + y) e
x dx dy
= 10 [(x + y − 1) e
x ]y0 dy =
1
0 [(2y − 1) e
y − (y − 1)] dy
= 2yey − 3ey − 12 y
2 + y 10 =
1
2 (7 − 2e)
 9. 1
0
x 2
0
x +2y
0 y dz dy dx =
1
0
x2
0 yx + 2y
2 dy dx
= 10
1
2 xy
2 + 23 y
3 x 2
0 dx =
1
0
1
2 x
5 + 23 x
6 dx
= 112 x
6 + 221 x
7 1
0 =
5
28
 10. Aqui E é a região que está abaixo do plano 3x + 2y + z = 6 e 
acima da região no plano xy limitada pelas retas x = 0, y = 0 e 
3x + 2y = 6, então 
 
E x dV =
2
0
3− 3x/2
0
6− 3x − 2y
0 x dz dy dx
= 20
3− 3x/2
0 6x − 3x
2 − 2xy dy dx
= 20 6x − 3x
2 3 − 32 x − x 3 −
3
2 x
2 dx
= 9 20 x − x
2 + 14 x
3 dx
= 9 12 x
2 − 13 x
3 + 116 x
4 2
0 = 3
 11. 
 Por simetria, E z dV = 2 E z dV , onde E′ é a parte de 
E à esquerda [tendo como ponto de vista (10, 10, 0)] do plano 
x = y. Logo 
 
E z dV =
1
0
1
y
1− x
0 2z dz dx dy =
1
0
1
y (1 − x )
2 dx dy
= 10 −
1
3 (1 − x )
3 x =1
x = y dy =
1
0
1
3 (1 − y)
3 dy
= 112 (1 − y)
4 1
0 =
1
12
 12. O plano 2x + 3y + 6z = 12 intersecta o plano xy quando 
 
2x + 3y + 6(0) = 12 ⇒ y = 4 − 23 x . Logo
E = (x, y, z ) | 0 ≤ x ≤ 6, 0 ≤ y ≤ 4 − 23 x ,
0 ≤ z ≤ 16 (12 − 2x − 3y)
e
V = 60
4− 2x/3
0
(12− 2x − 3y ) / 6
0 dz dy dx
= 16
6
0
4− 2x/3
0 (12 − 2x − 3y) dy dx
= 16
6
0 12y − 2xy −
3
2 y
2 y =4 − 2x/3
y =0 dx
=
1
6
6
0
(12 − 2x )2
3
−
3
2
12 − 2x
9
dx
= 136
6
0 (12 − 2x )
2 dx = 136 −
1
6 (12 − 2x )
3 6
0 = 8
15.7 SOLUÇÕES Revisão técnica: Ricardo Miranda Martins – IMECC – Unicamp
4  SEÇÃO 15.7 INTEGRAIS TRIPLAS
 13. 
V =
1
− 1
4− 4x 2
− 4− 4x 2
z +2
0
dy dz dx
= 2
1
0
4− 4x 2
− 4− 4x 2
z +2
0
dy dz dx
= 2
1
0
4− 4x 2
− 4− 4x 2
(z + 2) dz dx
= 2
1
0
1
2
z 2 + 2z
z =2 1− x 2
z = − 2 1− x 2
dx
= 2 10 8 1 − x 2 dx
= 16 12 x 1 − x 2 +
1
2 sen
− 1 x 10 = 4pi
 14. 
V = 10
x
− x
1− x
0 dz dy dx =
1
0
x
− x (1 − x ) dy dx
= 10 2 x (1 − x ) dx =
1
0 2 x − x
3 / 2 dx
= 2 23 x
3 / 2 − 25 x
5 / 2
1
0
= 2 23 −
2
5 =
8
15
x
 15. Os paraboloides z = x 2 + y2 e z = 18 − x 2 − y2
intersectam-se quando x 2 + y2 = 18 − x 2 − y2 ⇒
2x 2 + 2y2 = 18 ⇒ x 2 + y2 = 9. Logo,
E = (x, y, z ) | x 2 + y2 ≤ 9,
x 2 + y2 ≤ z ≤ 18 − x 2 − y2
Seja D = (x, y) | x 2 + y2 ≤ 9 . Então
V = E dV = D
18− x 2− y 2
x 2+ y 2 dz dA
= D 18 − 2x
2 − 2y2 dA
= 2pi0
3
0 18 − 2r
2 r dr dθ
= 2pi0 9r
2 − 12 r
4 r =3
r =0 dθ =
2pi
0
81
2 dθ = 81pi
 
 16. 
2
0
sen x
− 1
z + x
z − x e
3x (5y + 2 z ) dy dz dx
= e6 − 3518 −
511
338 cos 4 −
140
169 sen 4 −
749
1521 .
No Maple, o comando é
int(int(int(f,y=z-x..z+x),z=-1..sin(x)),
x=0..2);.
 17. (a) m = 1− 1
1− y 2
− 1− y 2
4
4y 2 +4z2 x
2 + y2 + z 2 dx dz dy
(b) (x, y, z ) onde
x = 1m
1
− 1
1− y 2
− 1− y 2
4
4y 2 +4z2 x x
2 + y2 + z 2 dx dz dy
y = 1m
1
− 1
1− y 2
− 1− y 2
4
4y 2 +4z2 y x
2 + y2 + z 2 dx dz dy
z = 1m
1
− 1
1− y 2
− 1− y 2
4
4y 2 +4z2 z x
2 + y2 + z 2 dx dz dy
(c)
I z = 1− 1
1− y 2
− 1− y 2
4
4y 2 +4 z2 x
2+ y2 x 2 + y2 + z 2 dx dz dy
 18. V (E ) = (1)(1)(1)6 =
1
6 . A equação do plano através dos
últimos três vértices é x +y +z = 1, então
f med = 11 / 6
1
0
1− x
0
1 − x − y
0 (x + y + z ) dz dy dx
= 6 10
1− x
0 (x + y) ( 1 − x − y)+
1
2 (1 − x − y)
2 dy dx
= 3 10
1− x
0 1 − 2xy − x
2 − y2 dy dx
= 3 10
1− x
0 1 − (x + y)
2 dy dx
= 3 10 y −
1
3 (x + y)
3 y =1− x
y =0 dx
= 3 10 1 − x −
1
3 +
1
3 x
3 dx = 10 x
3 − 3x + 2 dx
= 14 −
3
2 + 2 =
3
4