Solução_Calculo_Stewart_6e

F.
292 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
59. Let A(S 1) be the surface area of that portion of the surface which lies above the plane z =0.ThenA(S) =2A(S 1).
Following Example 10, a parametric representation of S 1 is x = a sin φ cos θ, y = a sin φ sin θ,
z = a cos φ and |r φ × r θ | = a 2 sin φ. ForD, 0 ≤ φ ≤ π 2 and for each fixed φ, x − 1 2 a2 + y 2 ≤ 1
2 a2 or
a sin φ cos θ −
1
2 a2 + a 2 sin 2 φ sin 2 θ ≤ (a/2) 2 implies a 2 sin 2 φ − a 2 sin φ cos θ ≤ 0 or
sin φ (sin φ − cos θ) ≤ 0. But0 ≤ φ ≤ π 2 ,socos θ ≥ sin φ or sin π
2 + θ ≥ sin φ or φ − π 2 ≤ θ ≤ π 2 − φ.
Hence D = (φ, θ) | 0 ≤ φ ≤ π 2 , φ − π 2 ≤ θ ≤ π 2 − φ .Then
Thus A(S) =2a 2 (π − 2).
A(S 1 )= π/2 (π/2) − φ
0 φ − (π/2) a2 sin φdθdφ= a 2 π/2
(π − 2φ)sinφdφ
0
= a 2 [(−π cos φ) − 2(−φ cos φ +sinφ)] π/2
0
= a 2 (π − 2)
Alternate solution: Working on S 1 we could parametrize the portion of the sphere by x = x, y = y, z = a 2 − x 2 − y 2 .
x 2
Then |r x × r y| = 1+
a 2 − x 2 − y + y 2
2 a 2 − x 2 − y = a
2 a2 − x 2 − y and
2
A(S 1 )=
0 ≤ (x − (a/2)) 2 + y 2 ≤ (a/2) 2 a
= π/2
−π/2
Thus A(S) =4a 2 π
2 − 1 =2a 2 (π − 2).
Notes:
−a(a2 − r 2 ) 1/2 r = a cos θ
a2 − x 2 − y 2 dA = π/2
r =0
−π/2
a cos θ
0
a
√
a2 − r 2 rdrdθ
dθ = π/2
−π/2 a2 [1 − (1 − cos 2 θ) 1/2 ] dθ
= π/2
−π/2 a2 (1 − |sin θ|) dθ =2a 2 π/2
0
(1 − sin θ) dθ =2a 2 π
2 − 1
(1) Perhaps working in spherical coordinates is the most obvious approach here. However, you must be careful
in setting up D.
(2) In the alternate solution, you can avoid having to use |sin θ| by working in the first octant and then
multiplying by 4. However, if you set up S 1 as above and arrived at A(S 1)=a 2 π, you now see your error.
17.7 Surface Integrals ET 16.7
1. The faces of the box in the planes x =0and x =2have surface area 24 and centers (0, 2, 3), (2, 2, 3). The faces in y =0and
y =4have surface area 12 and centers (1, 0, 3), (1, 4, 3), and the faces in z =0and z =6have area 8 and centers (1, 2, 0),
(1, 2, 6). For each face we take the point Pij ∗ to be the center of the face and f(x, y, z) =e −0.1(x+y+z) ,sobyDefinition 1,
f(x, y, z) dS ≈ [f(0, 2, 3)](24) + [f(2, 2, 3)](24) + [f(1, 0, 3)](12)
S
+[f(1, 4, 3)](12) + [f(1, 2, 0)](8) + [f(1, 2, 6)](8)
=24(e −0.5 + e −0.7 ) + 12(e −0.4 + e −0.8 )+8(e −0.3 + e −0.9 ) ≈ 49.09