Solução_Calculo_Stewart_6e

F.
SECTION 16.3 TX.10 DOUBLE INTEGRALS OVER GENERAL REGIONS ET SECTION 15.3 ¤ 231
47. 4
0
2
√ x
1
y 3 +1 dy dx = 2
0
2
=
0
y
2
0
1
dx dy
y 3 +1
1 x=y
2
x
2
dy = y 3 +1
x=0
0
y 2
y 3 +1 dy
= 1 ln y 3 +1 2
= 1 (ln 9 − ln 1) = 1 ln 9
3 3 3
0
49.
1
π/2
0
arcsin y
= π/2
0
cos x 1+cos 2 xdxdy
sin x
0
cos x √ 1+cos 2 xdydx
= π/2
0
cos x √ 1+cos 2 x y y=sin x
y=0
= π/2
0
cos x √ 1+cos 2 x sin xdx
dx
= 0
−u √
1+u
1 2 du = − 1 3 1+u
2 3/2
0
1
√ √
= 1 3 8 − 1 =
1
3 2 2 − 1
Let u =cosx, du = − sin xdx,
dx = du/(− sin x)
51. D = {(x, y) | 0 ≤ x ≤ 1, − x +1≤ y ≤ 1} ∪ {(x, y) | −1 ≤ x ≤ 0, x +1≤ y ≤ 1}
D
∪ {(x, y) | 0 ≤ x ≤ 1, − 1 ≤ y ≤ x − 1} ∪ {(x, y) | −1 ≤ x ≤ 0, − 1 ≤ y ≤−x − 1}, all type I.
x 2 dA =
=4
1
1
0 1 − x
1
1
0
1 − x
0
1
x 2 dy dx + x 2 dy dx +
−1 x +1
1
x − 1
0
−1
0
x 2 dy dx +
−1
−x − 1
x 2 dy dx [by symmetry of the regions and because f(x, y) =x 2 ≥ 0]
−1
x 2 dy dx
=4 1
0 x3 dx =4 1
1
4 0
53. Here Q = (x, y) | x 2 + y 2 ≤ 1 ,x≥ 0,y ≥ 0 ,and0 ≤ (x 2 + y 2 ) 2 ≤
1 2
4 4
⇒ − 1
16 ≤−(x2 + y 2 ) 2 ≤ 0 so
e −1/16 ≤ e −(x2 +y 2 ) 2 ≤ e 0 =1since e t is an increasing function. We have A(Q) = 1 π
1 2
= π , so by Property 11,
4 2 16
e −1/16 A(Q) ≤ Q e−(x2 +y 2 ) 2 dA ≤ 1 · A(Q) ⇒ π 16 e−1/16 ≤ Q e−(x2 +y 2 ) 2 dA ≤ π or we can say
16
0.1844 < Q e−(x2 +y 2 ) 2 dA < 0.1964. (We have rounded the lower bound down and the upper bound up to preserve the
inequalities.)
55. The average value of a function f of two variables defined on a rectangle R was
defined in Section 16.1 [ET 15.1] as f ave = 1 f(x, y)dA. Extending
A(R) R
this definition to general regions D,wehavef ave = 1 f(x, y)dA.
A(D)
Here D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 3x},soA(D) = 1 (1)(3) = 3 and
2 2
f ave = 1
1 1
3x
f(x, y)dA = xy dy dx
A(D)
D 3/2 0 0
= 2 1
1 xy2 y=3x
dx = 1 1
3 0 2 y=0 3 0 9x3 dx = 3 x4 1
= 3
4 0 4
57. Since m ≤ f(x, y) ≤ M, mdA≤ f(x, y) dA ≤ MdAby (8) ⇒
D D D
m 1 dA ≤ f(x, y) dA ≤ M 1 dA by (7) ⇒ mA(D) ≤ f(x, y) dA ≤ MA(D) by (10).
D D D D
D