Solução_Calculo_Stewart_6e

F.
296 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
31. z = xy ⇒ ∂z/∂x = y, ∂z/∂y = x, so by Formula 4, a CAS gives
S xyz dS = 1 1 xy(xy) y
0 0 2 + x 2 +1dx dy ≈ 0.1642.
33. We use Formula 4 with z =3− 2x 2 − y 2 ⇒ ∂z/∂x = −4x, ∂z/∂y = −2y. The boundaries of the region
3 − 2x 2 − y 2 3
≥ 0 are − ≤ x ≤ 3
and −√ 3 − 2x
2 2 2 ≤ y ≤ √ 3 − 2x 2 ,soweuseaCAS(withprecisionreducedto
seven or fewer digits; otherwise the calculation may take a long time) to calculate
S
√
x 2 y 2 z 2 3/2
√ 3 − 2x 2
dS = √ x
− 3/2 −
√3 2 y 2 (3 − 2x 2 − y 2 ) 2 16x 2 +4y 2 +1dy dx ≈ 3.4895
− 2x 2
35. If S is given by y = h(x, z), thenS is also the level surface f(x, y, z) =y − h(x, z) =0.
∇f(x, y, z)
n =
|∇f(x, y, z)| = −h x i + j − h z k
√ ,and−n is the unit normal that points to the left. Now we proceed as in the
h
2 x +1+h 2 z
derivation of (10), using Formula 4 to evaluate
S
∂h
F · dS = F · n dS = (P i + Q j + R k) ∂x i − j + ∂h
∂z k ∂h
∂h
S
D
2 2 ∂h
∂x
+1+
∂x
∂z
2
+1+
where D is the projection of S onto the xz-plane. Therefore F · dS = P ∂h
S
D ∂x − Q + R ∂h
dA.
∂z
37. m = S KdS = K · 4π 1
2 a2 =2πa 2 K;bysymmetryM xz = M yz =0,and
M xy = zK dS = K 2π π/2
(a cos φ)(a 2 sin φ) dφ dθ =2πKa 3 − 1 cos 2φ π/2
S 0 0 4
= πKa 3 .
0
Hence (x, y, z) = 0, 0, 1 2 a .
39. (a) I z = S (x2 + y 2 )ρ(x, y, z) dS
(b) I z = S (x2 + y 2 )
10 −
x 2 + y 2 dS =
= 2π
0
4
1
√
2(10r 3 − r 4 ) dr dθ =2 √ 2 π 4329
10
1 ≤ x 2 + y 2 ≤ 16
=
4329
5
(x 2 + y 2 ) 10 − √2
x 2 + y 2 dA
√
2 π
2 ∂h
dA
∂z
41. Therateofflow through the cylinder is the flux ρv · n dS = ρv · dS. We use the parametric representation
S S
r(u, v) =2cosu i +2sinu j + v k for S,where0 ≤ u ≤ 2π, 0 ≤ v ≤ 1,sor u = −2sinu i +2cosu j, r v = k,andthe
outward orientation is given by r u × r v =2cosu i +2sinu j. Then
S ρv · dS = ρ 2π
1
0 0 v i +4sin 2 u j +4cos 2 u k · (2 cos u i +2sinu j) dv du
= ρ 2π
0
1
0
= ρ sin u +8 − 1 3
2v cos u +8sin 3 u dv du = ρ 2π
0 cos u +8sin 3 u du
(2 + sin 2 u)cosu 2π
0
=0kg/s