Solução_Calculo_Stewart_6e

F.
226 ¤ CHAPTER 16 MULTIPLE INTEGRALS ET CHAPTER 15
TX.10
11.
13.
1
0
2
0
1 (u − 0 v)5 du dv = 1
1 (u − v)6 u=1
dv = 1 1
0 6 u=0 6 0 (1 − v) 6 − (0 − v) 6 dv
= 1 1
6 0 (1 − v) 6 − v 6
dv = 1 6 −
1
(1 − 7 v)7 − 1 v7 1
7 0
= − 1 [(0 + 1) − (1 + 0)] = 0
42
π r 0 sin2 θdθdr = 2
rdr π
0 0 sin2 θdθ [as in Example 5] = 2
rdr π
0 0
= 1
2 r2 2
0· 1
2
θ −
1
2 sin 2θ π
0
=2· 1
2
[(π − 0) − (0 − 0)] = π
1
(1 − cos 2θ) dθ
2
1
=(2− 0) ·
2 π −
1
sin 2π 2
− 0 − 1 2
sin 0
15.
17.
19.
21.
R (6x2 y 3 − 5y 4 ) dA = 3
0
R
xy 2 1
x 2 +1 dA =
= 1 2
0
1
0 (6x2 y 3 − 5y 4 ) dy dx = 3
3
0 2 x2 y 4 − y 5 y=1
dx = 3
3
y=0 0 2 x2 − 1 dx
= 1
2 x3 − x 3
0 = 27 2 − 3= 21 2
3
−3
π/6
π/3
x sin(x + y) dy dx
0 0
= π/6
0
xy 2 1
x 2 +1 dy dx =
1
(ln 2 − ln 1) · (27 + 27) = 9 ln 2
3
−x cos(x + y)
y = π/3
= x sin x − sin x + π 3
y =0
π/6
− π/6
0 0
= π 6
1
2 − 1 − − cos x +cos x + π 3
R xyex2y dA= 2
0
1
0 xyex2y dx dy = 2
0
0
x
3
x 2 +1 dx
−3
dx = π/6
0 x cos x − x cos x +
π
3 dx
sin x − sin x +
π
π/6
= 1 2 [(e2 − 2) − (1 − 0)] = 1 2 (e2 − 3)
0
1 3 1 1
y 2 dy =
2 ln(x2 +1)
0
3 y3 −3
3
dx [by integrating by parts separately for each term]
= − π − − √ 3
+0− −1+ 1 = √ 3−1
− π 12 2 2
2 12
x=1
1
2 ex2 y
dy = 1 2
2 0 (ey − 1) dy = 1 2 e y − y 2
x=0
0
23. z = f(x, y) =4− x − 2y ≥ 0 for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Sothesolid
is the region in the first octant which lies below the plane z =4− x − 2y
and above [0, 1] × [0, 1].
25. V = (12 − 3x − 2y) dA = 3
1 (12 − 3x − 2y) dx dy = 3
R −2 0 −2 12x −
3
2 x2 − 2xy x=1
= 3
21 − 2y dy = 21
y − y2 3
= 95
−2 2 2 −2 2
27. V = 2
1
−2 −1 1 −
1
4 x2 − 1 y2 dx dy =4 2 1
9 0 0 1 −
1
4 x2 − 1 y2 dx dy
9
=4 2
0
x −
1
12 x3 − 1 9 y2 x x =1
dy =4 2
x =0 0
x=0 dy
11
− 1 y2 dy =4 11
y − 1 y3 2 83
=4· = 166
12 9 12 27 0 54 27