Solução_Calculo_Stewart_6e

F.
51. (a) A(S) = 1+
D
2 ∂z
+
∂x
∂z
∂y
Using the Midpoint Rule with f(x, y) =
A(S) ≈
3
i =1j =1
(b) Using a CAS we have A(S) =
to the first decimal place.
53. z =1+2x +3y +4y 2 ,so
A(S) = 1+
D
Using a CAS, we have
4
1
1
0
or 45
8
SECTIONTX.10
17.6 PARAMETRIC SURFACES AND THEIR AREAS ET SECTION 16.6 ¤ 291
2
dA =
6
0
4
0
1+ 4x2 +4y 2
dy dx.
(1 + x 2 + y 2 )
4
1+ 4x2 +4y 2
, m =3, n =2we have
(1 + x 2 + y 2 )
4
2
f
x i , y j ∆A =4[f(1, 1) + f(1, 3) + f(3, 1) + f(3, 3) + f(5, 1) + f(5, 3)] ≈ 24.2055
2 ∂z
+
∂x
6
4
0
∂z
∂y
14 + 48y +64y2 dy dx = 45
8
√
14 +
15
ln 11 √ 5+3 √ 70
16 3 √ 5+ √ . 70
0
2
dA =
4
1+ 4x2 +4y 2
dy dx ≈ 24.2476. This agrees with the estimate in part (a)
(1 + x 2 + y 2 )
4
1
1
0
1+4+(3+8y)2 dy dx =
4
1
1
√
14 +
15
16 ln 11 √ 5+3 √ 14 √ 5 − 15
16 ln 3 √ 5+ √ 14 √ 5
0
14 + 48y +64y2 dy dx.
55. (a) x = a sin u cos v, y = b sin u sin v, z = c cos u ⇒
x 2
a + y2
2 b + z2
2 c =(sinu cos 2 v)2 +(sinu sin v) 2 +(cosu) 2
=sin 2 u +cos 2 u =1
and since the ranges of u and v are sufficient to generate the entire graph,
the parametric equations represent an ellipsoid.
(b)
(c) From the parametric equations (with a =1, b =2,andc =3), we calculate
r u =cosu cos v i +2cosu sin v j − 3sinu k and r v = − sin u sin v i +2sinu cos v j. So
r u × r v =6sin 2 u cos v i +3sin 2 u sin v j +2sinu cos u k, and the surface area is given by
A(S) = 2π
0
π
0
|ru × rv| du dv = 2π
0
π
0
36 sin 4 u cos 2 v +9sin 4 u sin 2 v +4cos 2 u sin 2 ududv
57. To find the region D: z = x 2 + y 2 implies z + z 2 =4z or z 2 − 3z =0.Thusz =0or z =3are the planes where the
surfaces intersect. But x 2 + y 2 + z 2 =4z implies x 2 + y 2 +(z − 2) 2 =4,soz =3intersects the upper hemisphere.
Thus (z − 2) 2 =4− x 2 − y 2 or z =2+ 4 − x 2 − y 2 . Therefore D is the region inside the circle x 2 + y 2 +(3− 2) 2 =4,
that is, D = (x, y) | x 2 + y 2 ≤ 3 .
A(S) = 1+[(−x)(4 − x 2 − y 2 ) −1/2 ] 2 +[(−y)(4 − x 2 − y 2 ) −1/2 ] 2 dA
D
2π
√
3
2π
√
=
1+ r2
3
4 − r rdrdθ= 2rdr
2π
√
−2(4 dθ = − r 2 ) 1/2 r= √ 3
2 4 − r
2 r=0
0
0
= 2π
0 (−2+4)dθ =2θ 2π
0
=4π
0
0
0
dθ