Solução_Calculo_Stewart_6e

F.
378 ¤ CHAPTER 8 PROBLEMS PLUS
TX.10
(c) The area on the sphere lies between planes y = y 1 and y = y 2,wherey 2 − y 1 = h. Thus, we compute the surface area on
y2
2 dx
y2
the sphere to be S = 2πx 1+ dy = 2πr dy =2πr(y 2 − y 1)=2πrh.
y 1
dy
y 1
This equals the lateral area of a cylinder of radius r and height h, since such
a cylinder is obtained by rotating the line x = r about the y-axis, so the
surface area of the cylinder between the planes y = y 1 and y = y 2 is
y2
A = 2πx
y 1
=2πry
y 2
1+
2 dx
y2
dy = 2πr 1+0
dy
2 dy
y 1
y=y 1
=2πr(y 2 − y 1)=2πrh
(d) h =2r sin 23.45 ◦ ≈ 3152 mi, so the surface area of the
Torrid Zone is 2πrh ≈ 2π(3960)(3152) ≈ 7.84 × 10 7 mi 2 .
5. (a) Choose a vertical x-axis pointing downward with its origin at the surface. In order to calculate the pressure at depth z,
consider n subintervals of the interval [0,z] by points x i and choose a point x ∗ i ∈ [x i−1 ,x i ] for each i. The thin layer of
water lying between depth x i−1 and depth x i has a density of approximately ρ(x ∗ i ), so the weight of a piece of that layer
with unit cross-sectional area is ρ(x ∗ i )g ∆x. The total weight of a column of water extending from the surface to depth z
(with unit cross-sectional area) would be approximately n ρ(x ∗ i )g ∆x. The estimate becomes exact if we take the limit
as n →∞;weight(orforce)perunitareaatdepthz is W = lim
i=1
n
n→∞ i=1
ρ(x ∗ i )g ∆x. Inotherwords,P (z) = z
0 ρ(x)gdx.
More generally, if we make no assumptions about the location of the origin, then P (z) =P 0 + z
ρ(x)gdx,whereP0 is
0
the pressure at x =0. Differentiating, we get dP/dz = ρ(z)g.
(b)
F = r
P (L + x) · 2 √ r
−r 2 − x 2 dx
=
r
P
−r 0 +
L+x
ρ
0 0 e z/H gdz · 2 √ r 2 − x 2 dx
= P 0
r
−r 2 √ r 2 − x 2 dx + ρ 0 gH r
−r
e (L+x)/H − 1 · 2 √ r 2 − x 2 dx
=(P 0 − ρ 0 gH) r
2 √ r
−r 2 − x 2 dx + ρ 0 gH r
−r e(L+x)/H · 2 √ r 2 − x 2 dx
=(P 0 − ρ 0 gH) πr 2 + ρ 0 gHe L/H r
−r ex/H · 2 √ r 2 − x 2 dx