Solução_Calculo_Stewart_6e

F.
364 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION
TX.10
7. Set up a vertical x-axis as shown. Then the area of the ith rectangular strip is
2 − √ 2
√
x ∗ i ∆x. By similar triangles, wi 3 − x
∗
3 2 = √3 i
,sow i =2− √ 2
x ∗ i .
3
The pressure on the strip is ρgx ∗ i , so the hydrostatic force on the strip is
ρgx ∗ i 2 − √ 2
x ∗
i ∆x and the hydrostatic force on the plate ≈ n
3
The total force
F = lim
n→∞ i=1
n
ρgx ∗ i
2 − √ 2
x ∗ i ∆x =
3
√ 3
0
i=1
ρgx ∗ i
2 − √ 2
x ∗ i ∆x.
3
ρgx 2 − √ 2
x dx = ρg
3
√ 3
= ρg x 2 − 2 √ 3
3 √ 3 x3 = ρg [(3 − 2) − 0] = ρg ≈ 1000 · 9.8 =9.8 × 10 3 N
0
0
2x − 2 √
3
x 2
dx
9. Set up coordinate axes as shown in the figure. The length of the ith strip is
2 25 − (yi ∗)2 and its area is 2 25 − (yi ∗)2 ∆y. The pressure on this strip is
approximately δd i =62.5(7 − yi ∗ ) andsotheforceonthestripisapproximately
62.5(7 − yi ∗ )2 25 − (yi ∗)2 ∆y. The total force
F = lim
n
n→∞ i=1
62.5(7 − y ∗ i )2 25 − (y ∗ i )2 ∆y =125 5
0 (7 − y) 25 − y 2 dy
5
=125 7 25 − y
0 2 dy − 5
y 25 − y
0 2 dy
=125 7 5
5
0 25 − y2 dy − − 1 (25 − 3 y2 ) 3/2 0
=125 7 1
π · 52 + 1 (0 − 125) =125
175π
− 125
4 3 4 3 ≈ 11,972 ≈ 1.2 × 10 4 lb
11. Set up a vertical x-axis as shown. Then the area of the ith rectangular strip is
a
h (2h − w i
x∗ i ) ∆x. By similar triangles, = 2a
2h − x ∗ i
2h ,sowi = a
h (2h − x∗ i ).
The pressure on the strip is δx ∗ i , so the hydrostatic force on the plate
≈ n
δx ∗ i
i=1
a
h (2h − x∗ i ) ∆x. The total force
F = lim
n
δx ∗ i
n→∞ i=1
= aδ
hx 2 − 1
h
x3 h
3
a
h (2h − x∗ i ) ∆x = δ a h aδ
x(2h − x) dx =
h
0
h
2h
3
0 = aδ
h
h 3 − 1 3 h3 = aδ
h
3
= 2 3 δah2
h
0
2hx − x
2 dx