Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 16 REVIEW ET CHAPTER 15 ¤ 263
43. (a) f(x, y) is a joint density function, so we know that R 2 f(x, y) dA =1.Sincef(x, y) =0outside the rectangle
[0, 3] × [0, 2], we can say
R 2 f(x, y) dA = ∞
−∞
Then 15C =1 ⇒ C = 1 15 .
(b) P (X ≤ 2,Y ≥ 1) = 2
−∞
= 1 2
15 0
= C 3
0
∞ f(x, y) dy dx = 3 2
C(x + y) dy dx
−∞ 0 0
xy +
1
2 y2 y=2
y=0 dx = C 3
0 (2x +2)dx = C x 2 +2x 3
0 =15C
∞
f(x, y) dy dx = 2 2
1 0 1
x +
3
2 dx =
1 1
15 2 x2 + 3 x 2
= 1
2 0 3
1
(x, y) dy dx = 1 2
15 15 0 xy +
1
y2 y=2
dx
2 y=1
(c) P (X + Y ≤ 1) = P ((X, Y ) ∈ D) where D is the triangular region shown in
the figure. Thus
1−x
P (X + Y ≤ 1) = f(x, y) dA = 1
D 0 0
= 1 1
15 0 xy +
1
y2 y=1−x
2
= 1 1
15 0
= 1
30
y=0
dx
1
(x + y) dy dx
15
x(1 − x)+
1
(1 − x)2 dx
2
1 (1 −
0 x2 ) dx = 1
30 x −
1
x3 1
= 1
3 0 45
45.
1
1
1−y
f(x, y, z) dz dy dx = 1
−1 x 2 0 0
1−z
0
√ y
− √ f(x, y, z) dx dy dz
y
47. Since u = x − y and v = x + y, x = 1 (u + v) and y = 1 2 2
(v − u).
∂(x, y)
Thus
∂(u, v) = 1/2 1/2
−1/2 1/2 = 1
2
R
and x − y
4
0
x + y dA = u 1 4
dv
du dv = − = − ln 2.
2 −2 v 2
2 v
49. Let u = y − x and v = y + x so x = y − u =(v − x) − u ⇒ x = 1 (v − u) and y = v − 1 (v − u) = 1 2 2 2
(v + u).
∂(x, y)
∂(u, v) = ∂x ∂y
∂u ∂v − ∂x
∂y
∂v ∂u =
− 1 1
2 2 −
1 1
= − 1 = 1 . R is the image under this transformation of the square
2 2 2 2
with vertices (u, v) =(0, 0), (−2, 0), (0, 2),and(−2, 2). So
R
xy dA =
2
0
0
−2
v 2 − u 2
4
1
du dv = 1 2
2
8 0 v 2 u − 1 u3 u=0
dv = 1 2
3 u=−2 8 0
2v 2 − 8 3 dv =
1 2
8 3 v3 − 8 v 2
=0
3 0
This result could have been anticipated by symmetry, since the integrand is an odd function of y and R is symmetric about
the x-axis.
51. For each r such that D r lies within the domain, A(D r )=πr 2 , and by the Mean Value Theorem for Double Integrals there
exists (x r ,y r ) in D r such that f (x r ,y r )= 1
f(x, y) dA. But lim
πr (x r,y 2 r )=(a, b),
D r
r→0 +
1
so lim
f (x, y) dA = lim
r→0 + πr f(x r,y 2 r )=f(a, b) by the continuity of f.
D r
r→0 +