Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 8 REVIEW ¤ 375
1. y = 1 6 (x2 +4) 3/2 ⇒ dy/dx = 1 4 (x2 +4) 1/2 (2x) ⇒
1+(dy/dx) 2 =1+
Thus, L = 3
0
1
2 x(x2 +4) 1/2 2
=1+
1
4 x2 (x 2 +4)= 1 4 x4 + x 2 +1= 1
2 x2 +1 2 .
1
2 x2 +1 2 dx = 3
1
0 2 x2 +1 dx = 1
6 x3 + x 3
= 15 . 0 2
3. (a) y = x4
16 + 1
2x 2 = 1
16 x4 + 1 2 x−2 ⇒ dy
dx = 1 4 x3 − x −3
⇒
1+(dy/dx) 2 =1+ 1
4 x3 − x −32 =1+ 1 16 x6 − 1 + 2 x−6 = 1 16 x6 + 1 + 2 x−6 = 1
4 x3 + x −32 .
Thus, L = 2
1
1 4 x3 + x −3 dx = 1
16 x4 − 1 x−2 2
=
1 − 1 2 1 8 − 1 − 1
16 2 =
21
. 16
(b) S = 2
2πx 1
1 4 x3 + x −3 dx =2π 2
1
1 4 x4 + x −2 dx =2π
1
20 x5 − 1 2
x 1
=2π 32
20 − 1 2
−
1
− 1 20
=2π 8
− 1 − 1 +1 5 2 20
=2π 41
20
=
41
10 π
5. y = e −x2 ⇒ dy/dx = −2xe −x2 ⇒ 1+(dy/dx) 2 =1+4x 2 e −2x2 . Letf(x) = 1+4x 2 e −2x2 .Then
L =
3
0
f(x) dx ≈ S 6 =
(3 − 0)/6
3
[f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + f(3)] ≈ 3.292287
7. y = x
√ √ √
1 t − 1 dt ⇒ dy/dx = x − 1 ⇒ 1+(dy/dx) 2 =1+ x − 1 = √ x.
Thus, L = √
16
16
16
1 xdx= x 1/4 dx =
x 4 5/4 = 4 124
(32 − 1) = .
1 5
5 5
1
9. As in Example 1 of Section 8.3,
a
2 − x = 1 2
⇒ 2a =2− x and w =2(1.5+a) =3+2a =3+2− x =5− x.
Thus, F = 2
ρgx(5 − x) dx = ρg 5
0 2 x2 − 1 x3 2
= ρg
10 − 8 3 0 3 =
22
22
δ [ρg = δ] ≈ · 62.5 ≈ 458 lb.
3 3
11. A = 4
√
0 x −
1
x dx =
2
x = 1 4 x √ x − 1 x dx = 3 A 0 2 4
= 3 4
y = 1 A
4
2
3 x3/2 − 1 4 x2 = 16 − 4= 4 3 3
0
4
0
2
5 x5/2 − 1 6 x3 4
0
= 3 4
64
5 − 64 6
4
0
x 3/2 − 1 2 x2
dx
=
3
4
√ 2
1
2 x − 1 2
x
dx = 3 4
2 4 0
Thus, the centroid is (x, y) = 8
5 , 1 .
64
30 =
8
5
1
2 x −
1
x2
dx = 3 1
4 8 2 x2 − 1 x3 4
= 3
12 0 8
8 −
16
3 =
3 8
8 3 =1
13. An equation of the line passing through (0, 0) and (3, 2) is y = 2 x. A = 1 · 3 · 2=3. Therefore, using Equations 8.3.8,
3 2
x = 1 3 x 2
x
3 0 3
dx = 2 27 x
3 3
=2and y = 1 3
1 2
0 3 0 2 3 x2 dx = 2
81 x
3 3
= 2 . Thus, the centroid is (x, y) =
0 3
2, 2 3 .