Solução_Calculo_Stewart_6e

F.
240 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
center), the exposure of a person at A is
E = kf(P, A) dA = k
D
(b) If A =(0, 0),then
E = k
= k
D
2π
10
0
D
1 − 1
x2 + y
20
2 dx dy
0
=2πk 50 − 50 3
1 − r
20
20 − d(P, A)
20
r
2
10
rdrdθ=2πk
2 − r3
60
0
=
200
3
πk ≈ 209k
dA = k 1 −
D
(x − x0) 2 +(y − y 0) 2
dx dy
20
For A at the edge of the city, it is convenient to use a polar coordinate system centered at A. Then the polar equation for
the circular boundary of the city becomes r =20cosθ instead of r =10, and the distance from A to a point P in the city
is again r (see the figure). So
E = k
π/2
20 cos θ
−π/2
0
1 − r
π/2
r
2
r=20 cos θ
rdrdθ= k
20
−π/2 2 − r3
dθ
60
r=0
1 + 1 cos 2θ − 2 2 2 3
= k π/2
−π/2 200 cos 2 θ − 400
3 cos3 θ dθ =200k π/2
−π/2
1 − sin 2 θ cos θ dθ
=200k 1
θ + 1 sin 2θ − 2 sin θ + 2 · 1
2 4 3 3 3 sin3 θ π/2
=200k π
+0− 2 + 2 + π +0− 2 +
2
−π/2 4 3 9 4 3 9
=200k π
2 − 8 9
≈ 136k
Therefore the risk of infection is much lower at the edge of the city than in the middle, so it is better to live at the edge.
16.6 Triple Integrals ET 15.6
1.
3.
5.
B xyz2 dV = 1 3
0 0
1
z
0 0
3
1
0 0
= 1
0
x+z
6xz dy dx dz = 1
z
0 0 0
2
−1 xyz2 dy dz dx = 1 3
0 0
1
2 xy2 z 2 y=2
dz dx = 1 3
y=−1 0 0
1 xz3 z=3
dx = 1 27
27
xdx= x2 1
= 27
2 z=0 0 2 4 0 4
y=x+z
6xyz
= 1
0
√ 1−z 2
0
ze y dx dz dy = 3
3
2 xz2 dz dx
dx dz = 1
z
6xz(x + z) dx dz
y=0
0 0
2x 3 z +3x 2 z 2 x=z
dz = 1
x=0 0 (2z4 +3z 4 ) dz = 1
0 5z4 dz = z 5 1
√
= 3
0
xze
y x= 1−z 2
dz dy = 3
1 zey√ 1 − z
x=0
0 0 2 dz dy
z=1
− 1 (1 − 3 z2 ) 3/2 e y dy = 3
0
1
0 0
z=0
1
3 ey dy = 1 ey 3
= 1
3 0 3 (e3 − 1)
0 =1
7.
π/2
y
0 0
x cos(x + y + z)dz dx dy = π/2
y
z=x
0 0 0 sin(x + y + z) dx dy
z=0
= π/2
0
= π/2
0
= π/2
0
y
0
[sin(2x + y) − sin(x + y)] dx dy
−
1
2 cos(2x + y)+cos(x + y) x=y
x=0 dy
−
1
2 cos 3y +cos2y + 1 2 cos y − cos y dy
= − 1 6 sin 3y + 1 2 sin 2y − 1 2 sin y π/2
0
= 1 6 − 1 2 = − 1 3
9.
E 2xdV = 2
√ 4−y 2
0 0
x 2 y x=
= 2
0
y 2xdzdxdy = 2
√ 4−y 2
0
0
√
4−y 2
dy = 2
(4 − x=0 0 y2 )ydy= 2y 2 − 1 y4 2
4
0
z=y
2xz dx dy = 2
√ 4−y 2
2xy dx dy
z=0 0 0
0 =4