Solução_Calculo_Stewart_6e

F.
TX.10
PROBLEMS PLUS
1. x 2 + y 2 ≤ 4y ⇔ x 2 +(y − 2) 2 ≤ 4,soS is part of a circle, as shown
in the diagram. The area of S is
1
0
4y − y2 dy 113
=
y−2
2
4y − y2 +2cos −1 1
2−y
2
0
√
= − 1
2 3+2cos
−1 1
2 − 2cos −1 1
= − √ 3
+2
π
2 3 − 2(0) =
2π
− √ 3
3 2
[a =2]
Another method (without calculus): Note that θ = ∠CAB = π , so the area is
3
(area of sector OAB) − (area of 4ABC) = 1 2 2
2 π
− 1 (1)√ 3= 2π − √ 3
3 2 3 2
3. (a) The two spherical zones, whose surface areas we will call S 1 and S 2 ,are
generated by rotation about the y-axis of circular arcs, as indicated in the figure.
The arcs are the upper and lower portions of the circle x 2 + y 2 = r 2 that are
obtained when the circle is cut with the line y = d. The portion of the upper arc
in the first quadrant is sufficient to generate the upper spherical zone. That
portion of the arc can be described by the relation x = r 2 − y 2 for
d ≤ y ≤ r.Thus,dx/dy = −y/ r 2 − y 2 and
2
dx
ds = 1+ dy =
dy
From Formula 8.2.8 we have
r
2 dx
r
S 1 = 2πx 1+ dy =
dy
d
d
1+ y2
r 2 − y 2 dy =
r 2
r 2 − y 2 dy =
rdy
r2 − y 2
2π
rdy
r
r 2 − y 2
r2 − y = 2πr dy =2πr(r − d)
2
Similarly, we can compute S 2 = d
2πx 1+(dx/dy)
−r 2 dy = d
2πr dy =2πr(r + d). Note that S −r 1 + S 2 =4πr 2 ,
the surface area of the entire sphere.
d
(b) r =3960mi and d = r (sin 75 ◦ ) ≈ 3825 mi,
so the surface area of the Arctic Ocean is about
2πr(r−d) ≈ 2π(3960)(135) ≈ 3.36×10 6 mi 2 .
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