Solução_Calculo_Stewart_6e

F.
290 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
37. The surface S is given by z = f(x, y) =6− 3x − 2y which intersects the xy-plane in the line 3x +2y =6,soD is the
triangular region given by (x, y) 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 −
3
x 2
. By Formula 9, the surface area of S is
2 2 ∂z ∂z
A(S)= 1+ + dA
D ∂x ∂y
=
1+(−3)2 +(−2)
D
2 dA = √ 14 dA = √ 14 A(D) = √ 14 1 · 2 · 3 =3 √ 14.
D 2
39. z = f(x, y) = 2 3 (x3/2 + y 3/2 ) and D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 }. Thenf x = x 1/2 , f y = y 1/2 and
A(S)=
1+( √ x ) 2 + √ 2
D
y dA = 1
1 √
0 0 1+x + ydydx
=
y=1
1 2
(x + y
0 3 +1)3/2 dx = 2 1
(x +2) 3/2 3
− (x +1) 3/2 dx
0
y=0
1
= 2 2
(x 3 5 +2)5/2 − 2 (x 5 +1)5/2 = 4 15 (35/2 − 2 5/2 − 2 5/2 +1)= 4
15 (35/2 − 2 7/2 +1)
0
41. z = f(x, y) =xy with 0 ≤ x 2 + y 2 ≤ 1,sof x = y, f y = x ⇒
A(S)= D
= 2π
0
1+y2 + x 2 dA = 2π
0
1
3
2
√
2 − 1
dθ =
2π
3
1
0
√
r2 +1rdrdθ= 2π
0
2
√
2 − 1
r=1
1
3 (r2 +1) 3/2 dθ
r=0
43. z = f(x, y) =y 2 − x 2 with 1 ≤ x 2 + y 2 ≤ 4. Then
A(S)=
1+4x2 +4y
D
2 dA = 2π 2
√
1+4r2 rdrdθ= 2π
dθ 2
r √ 1+4r
0 1 0 1 2 dr
= θ
2π
2
1
(1 + √ √
0 12 4r2 ) 3/2 = π 6 17 17 − 5 5
1
45. A parametric representation of the surface is x = x, y =4x + z 2 , z = z with 0 ≤ x ≤ 1, 0 ≤ z ≤ 1.
Hence r x × r z =(i +4j) × (2z j + k) =4i − j +2z k.
Note: In general, if y = f(x, z) then r x × r z = ∂f
∂x i − j + ∂f
2 2 ∂f ∂f
∂z
D
k and A (S) = 1+ + dA. Then
∂x ∂z
A(S)= 1 1
√
0 0 17 + 4z2 dx dz = 1
√
0 17 + 4z2 dz
√
= 1 2 z 17 + 4z2 + 17 ln √
2 2z + 4z2 +17 1
= √ √ √
21
0 2
+ 17 4 ln 2+ 21 − ln 17
47. r u = h2u, v, 0i, r v = h0,u,vi,andr u × r v = v 2 , −2uv, 2u 2 .Then
A(S)= D |ru × rv| dA = 1 2
0 0
= 1
0
2
0 (v2 +2u 2 ) dv du = 1
0
√
v4 +4u 2 v 2 +4u 4 dv du = 1 2
0 0
(v2 +2u 2 ) 2 dv du
1
3 v3 +2u 2 v v=2
du = 1
8 +4u2 du = 8
u + 4 u3 1
=4
v=0 0 3 3 3 0
49. z = f(x, y) =e −x2 −y 2 with x 2 + y 2 ≤ 4.
A(S) =
1+ −2xe
D
−x2 −y 22 + −2ye −x2 −y 22 dA =
1+4(x2 + y
D 2 )e −2(x2 +y 2 )
dA
= 2π
2
1+4r2 e
0 0 −2r2 rdrdθ= 2π
dθ 2
r 1+4r
0 0 2 e −2r2 dr =2π 2
r 1+4r
0 2 e −2r2 dr ≈ 13.9783