Solução_Calculo_Stewart_6e

F.
SECTION TX.10 16.5 APPLICATIONS OF DOUBLE INTEGRALS ET SECTION 15.5 ¤ 235
37. (a) We integrate by parts with u = x and dv = xe −x2 dx. Thendu = dx and v = − 1 2 e−x2 ,so
∞
0
x 2 e −x2 t
dx = lim
t→∞ 0 x2 e −x2 dx = lim − 1 xe−x2 t
+
t 1
t→∞ 2 0 2 e−x2 dx
0
= lim
− 1 te−t2
+ 1 ∞
t→∞ 2 2
e −x2 dx =0+ 1 ∞
0 2
e −x2 dx [by l’Hospital’s Rule]
0
= 1 ∞
4 −∞ e−x2 dx [since e −x2 is an even function]
= 1 √
4 π [by Exercise 36(c)]
(b) Let u = √ x.Thenu 2 = x ⇒ dx =2udu ⇒
∞ √
0 xe −x t √
dx = lim
t→∞ 0 xe −x √ t
dx = lim ue −u2 2udu=2 ∞
u 2 e −u2 du =2 1
√
t→∞ 0 0 4 π
[by part(a)] = 1 2
√ π.
16.5 Applications of Double Integrals ET 15.5
1. Q = D σ(x, y) dA = 3
1
= 3
1
2 (2xy + 0 y2 ) dy dx = 3
1 xy 2 + 1 y3 y=2
dx
3 y=0
4x +
8
3 dx = 2x 2 + 8 x 3
3 1 =16+16 = 64 C
3 3
3. m = ρ(x, y) dA = 2
1
D 0 −1 xy2 dy dx = 2
xdx 1
0 −1 y2 dy = 1
x2 2 1 y3 1
=2· 2 = 4 ,
2 0 3 −1 3 3
x = 1 xρ(x, y) dA = 3 2
1
m D 4 0 −1 x2 y 2 dy dx = 3 2
4 0 x2 dx 1
−1 y2 dy = 3 1 x3 2 1 y3 1
= 3 · 8 · 2 = 4 ,
4 3 0 3 −1 4 3 3 3
y = 1 yρ(x, y) dA = 3 2
1
m D 4 0 −1 xy3 dy dx = 3 2 xdx 1
4 0 −1 y3 dy = 3 1 x2 2 1 y4 1
= 3 · 2 · 0=0.
4 2 0 4 −1 4
Hence, (x, y) = 4
3 , 0 .
5. m = 2 3−x
(x + y) dy dx = 2
0 x/2 0 xy +
1
y2 y=3−x
dx = 2
2 y=x/2 0 x 3 −
3
x 2
+ 1 (3 − 2 x)2 − 1 x2 dx
8
= 2
0 −
9
8 x2 + 9 2 dx = −
9 1 x3 + 9 x 2
=6,
8 3 2 0
M y = 2 3−x
0 x/2 (x2 + xy) dy dx = 2
0 x 2 y + 1 xy2 y=3−x
dx = 2
9 x − 9 x3 dx = 9 ,
2 y=x/2 0 2 8 2
M x = 2 3−y
(xy + 0 x/2 y2 ) dy dx = 2
1
0 2 xy2 + 1 y3 y=3−x
dx = 2
3 y=x/2 0 9 −
9
x 2
dx =9.
My
Hence m =6, (x, y) =
m , M
x 3
=
m 4 , 3 .
2
7. m = 1
e
x
ydydx= 1
1 y2 y=e x
dx = 1 1
0 0 0 2 y=0 2 0 e2x dx = 1 e2x 1
= 1
4 0 4 (e2 − 1),
M y = 1
e
x
xy dy dx = 1 1
0 0 2 0 xe2x dx = 1 1
2 2 xe2x − 1 e2x 1
= 1
4 0 8 (e2 +1),
M x = 1
e
x
y 2 dy dx = 1
1 y3 y=e x
dx = 1 1
0 0 0 3 y=0 3 0 e3x dx = 1 1 e3x 1
= 1
3 3 0 9 (e3 − 1).
1
Hence m = 1 4 (e2 8
− 1), (x, y) =
(e2 1
+1)
1
4 (e2 − 1) , 9 (e3 − 1)
1
4 (e2 − 1)
e 2 +1
=
2(e 2 − 1) , 4(e3 − 1)
.
9(e 2 − 1)