Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.7 SURFACE INTEGRALS ET SECTION 16.7 ¤ 293
3. We can use the xz-andyz-planes to divide H into four patches of equal size, each with surface area equal to 1 8
the surface
area of a sphere with radius √ 50,so∆S = 1 8 (4)π√ 50 2
=25π. Then(±3, ±4, 5) are sample points in the four patches,
and using a Riemann sum as in Definition 1, we have
f(x, y, z) dS ≈ f(3, 4, 5) ∆S + f(3, −4, 5) ∆S + f(−3, 4, 5) ∆S + f(−3, −4, 5) ∆S
5. z =1+2x +3y so ∂z ∂z
=2and
∂x
H
S x2 yz dS =
D
= √ 14 3
0
=(7+8+9+12)(25π) =900π ≈ 2827
∂y =3.ThenbyFormula4,
2 ∂z ∂z
+
∂x ∂y
x 2 yz
2
+1dA = 3
2
0 0 x2 y(1 + 2x +3y) √ 4+9+1dy dx
2
0 (x2 y +2x 3 y +3x 2 y 2 ) dy dx = √ 14 3
1
0 2 x2 y 2 + x 3 y 2 + x 2 y 3 y=2
dx y=0
= √ 14 3
0 (10x2 +4x 3 ) dx = √ 14 10
3 x3 + x 4 3
0 = 171 √ 14
7. S is the part of the plane z =1− x − y over the region D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}. Thus
S yz dS = y(1 − x − y) (−1)
D 2 +(−1) 2 +1dA = √ 3 1
1−x
0 0 y − xy − y
2
dy dx
= √ 3 1
1
0 2 y2 − 1 2 xy2 − 1 y3 y=1−x
3
dx = √ 3 1 1
(1 − y=0 0 6 x)3 dx = − √ 1
3
(1 − 24 x)4 = √ 3
24
0
9. r(u, v) =u 2 i + u sin v j + u cos v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ π/2, so
r u × r v =(2u i +sinv j +cosv k) × (u cos v j − u sin v k) =−u i +2u 2 sin v j +2u 2 cos v k and
|r u × r v| = u 2 +4u 4 sin 2 v +4u 4 cos 2 v = u 2 +4u 4 (sin 2 v +cos 2 v)=u √ 1+4u 2 (since u ≥ 0). Then by
Formula 2,
S yz dS = (u sin v)(u cos v) D |ru × rv| dA = π/2
1 (u sin v)(u cos v) · u √ 1+4u
0 0 2 du dv
= 1
√
0 u3 1+4u 2 du π/2
sin v cos vdv
0 let t =1+4u
2
⇒ u 2 = 1 (t − 1) and 1 dt = udu
4 8
= 5 1 · 1 (t − 1)√ tdt π/2
sin v cos vdv= 1 5
t 3/2 − √
t dt π/2
sin v cos vdv
1 8 4 0 32 1
0
= 1 32
5
2
5 t5/2 − 2 1
3 t3/2 2 sin2 v π/2
= 1
1
0 32
2
5 (5)5/2 − 2 3 (5)3/2 − 2 + 2 5 3
√
· 1
5
(1 − 0) =
2 48 5+
1
240
11. S is the portion of the cone z 2 = x 2 + y 2 for 1 ≤ z ≤ 3, or equivalently, S is the part of the surface z = x 2 + y 2 over the
region D = (x, y) | 1 ≤ x 2 + y 2 ≤ 9 .Thus
2
x 2 z 2 dS = x 2 (x 2 + y 2 )
x
+
S
D
x2 + y 2
= x 2 (x 2 + y 2 )
D
x 2 + y 2
x 2 + y 2 +1dA = D
y
x2 + y 2 2
+1dA
√
2 x 2 (x 2 + y 2 ) dA = √ 2
2π
0
3
1
(r cos θ) 2 (r 2 ) rdrdθ
= √ 2 2π
cos 2 θdθ 3
0 1 r5 dr = √ 2 1
θ + 1 sin 2θ 2π 1 r6 3
= √
2 4 0 6
2(π) · 1
1 6 (36 − 1) = 364 √ 2
π
3