Solução_Calculo_Stewart_6e

F.
31. A = π/4
0
(cos x − sin x) dx = sin x +cosx π/4
0
= √ 2 − 1.
x = A −1 π/4
0
x(cos x − sin x) dx
TX.10 SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING ¤ 367
= A −1 x(sin x +cosx)+cosx − sin x π/4
0
= A −1 √
1
π
π √ 4 2 − 1 =
4
2 − 1
√ .
2 − 1
[integration by parts]
y = A −1 π/4
0
1
2 (cos2 x − sin 2 x) dx = 1 π/4
cos 2xdx= 1
π/4
2A 0 4A sin 2x = 1
0
4A = 1
4 √ 2 − 1 .
π √
2 − 4
Thus, the centroid is (x, y) =
4 √ 2 − 1 , 1
4 √ 2 − 1 ≈ (0.27, 0.60).
33. From the figure we see that y =0.Now
A = 5
2 √
5
5 − xdx=2 − 2 (5 − 0 3 x)3/2 =2
0+ 2 · 3 53/2 0
so
= 20 3
√
5,
x = 1 5 x√ 5 − x − − √ 5 − x
dx = 1 5 2x √ 5 − xdx
A 0 A 0
0√5
2 5 − u 2
u(−2u) du
= 1 A
= 4 A
u = √ 5 − x, x =5− u 2 ,
u 2 =5− x, dx = −2udu
√ 5
u 2 (5 − u 2 ) du = 4 5
0 A 3 u3 − 1 5 u5√ 5
= 3
0 5 √ 25
√ √
5 3 5 − 5 5 =5− 3=2.
Thus, the centroid is (x, y) =(2, 0).
35. The line has equation y = 3 x. A = 1 (4)(3) = 6,som = ρA = 10(6) = 60.
4 2
M x = ρ 4
0
1 3
2 4 x2 dx =10 4
0
M y = ρ 4
x 3
x dx = 15
0 4 2
9
32 x2 dx = 45 1 x3 4
= 45
16 3 0 16
64
3 =60
4
0 x2 dx = 15 1 x3 4
= 15 64
2 3 0 2 3 = 160
x = M y
m = 160
60 = 8 3 and y = M x
m = 60
60 =1. Thus, the centroid is (x, y) = 8
3 , 1 .
37. A =
=
2
0
2
(2 x − x 2 x
) dx =
4
ln 2 − 8 3
2
ln 2 − x3
3
x = 1 x(2 x − x 2 ) dx = 1 A 0
A
= 1 x2
x
2
A ln 2 −
2x
(ln 2) 2 − x4
4
0
= 1 A
2
0
− 1
ln 2 = 3
ln 2 − 8 3 ≈ 1.661418.
2
0
(x2 x − x 3 ) dx
[use parts]
8
ln 2 − 4
(ln 2) 2 − 4+ 1
(ln 2) 2
= 1 A
8
ln 2 − 3
(ln 2) 2 − 4 ≈ 1 (1.297453) ≈ 0.781
A
[continued]