Solução_Calculo_Stewart_6e

F.
246 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
37. m = E ρ(x, y, z) dV = 1
0
= 1
0 2y +2xy + y
2 y= √ x
dx = 1
y=0 0
√ x
1+x+y
2 dz dy dx = 1
√ x
0 0 0
2 √ x +2x 3/2 + x
0
2(1 + x + y) dy dx
dx =
1
4
3 x3/2 + 4 5 x5/2 + 1 2 x2 = 79
30
0
M yz = xρ(x, y, z) dV = 1
√ x
1+x+y
2xdzdydx= 1
√ x
2x(1 + x + y) dy dx
E 0 0 0 0 0
= 1
0 2xy +2x 2 y + xy 2 y= √ x
dx =
1
1
y=0 0 (2x3/2 +2x 5/2 + x 2 4
) dx =
5 x5/2 + 4 7 x7/2 + 1 3 x3 = 179
105
0
M xz = E yρ(x, y, z) dV = 1
0
= 1
0
y 2 + xy 2 + 2 3 y3 y= √ x
y=0
M xy = E zρ(x, y, z) dV = 1
0
√ x
1+x+y
2ydzdydx= 1
√ x
0 0 0
dx = 1
0
x + x 2 + 2 3 x3/2
dx =
√ x
1+x+y
2zdzdydx= 1
0 0 0
2y(1 + x + y) dy dx
0
1
1
2 x2 + 1 3 x3 + 4 15 x5/2 = 11
10
0
√ x
0
z
2 z=1+x+y
= 1
√ x
(1 + 2x +2y +2xy + x 2 + y 2 ) dy dx = 1
0 0 0 y +2xy + y 2 + xy 2 + x 2 y + 1 y3 y= √ x
3
=
1 √x
0 +
7
3 x3/2 + x + x 2 + x 5/2 dx =
z=0
1
2
3 x3/2 + 14
15 x5/2 + 1 2 x2 + 1 3 x3 + 2 7 x7/2 = 571
210
0
Thus the mass is 79 and the center of mass is (x, y, z) = Myz
30
m , Mxz
m , Mxy 358
=
m 553 , 33
79 , 571
.
553
dy dx = 1
√ x
(1 + x + y) 2 dy dx
0 0
y=0
dx
39. m = a a
0 0
= a
0
M yz = a a
a
0 0 0
= a
0
a
0 (x2 + y 2 + z 2 ) dx dy dz = a a
1
0 0 3 x3 + xy 2 + xz 2 x=a
dy dz = a a
1
x=0 0 0 3 a3 + ay 2 + az 2 dy dz
1
3 a3 y + 1 3 ay3 + ayz 2 y=a
dz = a
2
y=0 0 3 a4 + a 2 z 2 dz = 2
3 a4 z + 1 3 a2 z 3 a
= 2 0 3 a5 + 1 3 a5 = a 5
x 3 + x(y 2 + z 2 ) dx dy dz = a
a
1
0 0 4 a4 + 1 2 a2 (y 2 + z 2 ) dy dz
1
4 a5 + 1 6 a5 + 1 2 a3 z 2 dz = 1 4 a6 + 1 3 a6 = 7 12 a6 = M xz = M xy by symmetry of E and ρ(x, y, z)
Hence (x, y, z) = 7
12 a, 7
12 a, 7 12 a .
41. I x = L L
L
0 0 0
k(y2 + z 2 ) dz dy dx = k L L
0 0
By symmetry, I x = I y = I z = 2 3 kL5 .
Ly 2 + 1 3 L3 dy dx = k L
0
2
3 L4 dx = 2 3 kL5 .
43. I z = E (x2 + y 2 ) ρ(x, y, z) dV =
h
0 k(x2 + y 2 ) dz dA =
k(x 2 + y 2 )hdA
x 2 +y 2 ≤a 2 x 2 +y 2 ≤a 2
= kh 2π a
0 0 (r2 ) rdrdθ= kh 2π
dθ a
0 0 r3 dr = kh(2π) 1
a
4 0 4 a4 = 1 2 πkha4
45. (a) m = 3
−3
√ √
9−x2
−
5−y
9−x 2 1
x2 + y 2 dz dy dx
Myz
(b) (x, y, z) =
m , Mxz
m , Mxy
where
m
M yz = 3
−3
M xy = 3
−3
(c) I z = 3
−3
√ √
9−x2
−
√ √
9−x2
−
√
√
9−x2
−
9−x 2 5−y
9−x 2 5−y
9−x 2 5−y
1
x x 2 + y 2 dz dy dx, M xz = 3
1
z x 2 + y 2 dz dy dx.
1
(x 2 + y 2 ) x 2 + y 2 dz dy dx = 3
−3
−3
√ 9−x
√
2
−
√ 9−x
√
2
−
9−x 2 5−y
9−x 2 5−y
1
y x 2 + y 2 dz dy dx, and
1
(x 2 + y 2 ) 3/2 dz dy dx