Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

326 D CHAPTER 16 VECTOR CALCULUS
53. z = f(x , y) = e- "' 2 -"2
with x
2
+ y 2 :=:; 4.
A(S) = IIo J 1 + ( -2xe->= 2 -Y 2 ) 2 + ( - 2ye-"' 2 - v 2 ) 2 dA = IIo y'1 + 4(x2 + y2)e-2(>= 2 +u 2 > dA
= I:1< I: v'1 + 4r 2 e - 2 r 2 r dr dB = I:" dB I: r y'1 + 4r2e- 2r 2 dr = 27r I: r V1 + 4r 2 e- 2r 2 dr ~ 13.9783
Using the Midpoint Rule with f(x, y) =
4x 2 + 4y 2
1 + ( 1 2 2 ) 4
. +x +Y
4x 2 + 4y 2
1 + (1 2 2)4 dydx.
+x +y
, m = 3, n = 2 we have
3 2
A(S) ~ 2: 2: f(xi, 'il; ) ~A = 4 (!(1, 1) + /(1, 3) + /(3, 1) + / (3, 3) + /(5, 1) + / (5, 3)] ~ 24.2055
i = lj = l
(b) Using a CAS we have A(S) = 1614
1 + (l 4 x 2 :
4 Y:)4 dy dx ~ 24.2476. This agrees with the estimate in part (a)
+x +y . -
to the first decimal place.
57. z = 1 + 2x + 3y + 4y 2 , so
Us ing a CAS, we have
or !!! v'I4 + ll Jn n ~+ V'J!
8 16 3 0 + 70 .
59. (a) x = a sin u cosv, y = bsin u sin v, z = ccos u => (b) ~~
2
IZfi
71.
z 0
and s ince the ranges of u and v ~ sufficient to generate the entire graph,
the parametric equations represent an ellipsoid.
(c) From the parametric equations (with a = 1, b = 2, and c = 3),
- 2
- 2
0
y
'lii!
~
-
we calculate r u = cosucosv i + 2cos usinvj - 3sin u k and
r, =-s in usin v i + 2sin ucos v j. So r,. x r, = 6 sin 2 t! cos v i + 3 sin 2 'usinvj + 2 sin u cos u k , and the surface
61. To find the region D : z = x 2 + y 2 implies z + z 2 = 4z or z 2 - 3z = 0. Thus z = 0 or z = 3 are the planes where the
surfaces intersect. But x 2 + y 2 + z 2 = 4z implies x 2 + y 2 + (z -
2) 2 = 4. so 'z = 3 intersects the upper hemisphere.
Thus (z - 2) 2 = 4 - x 2 - y 2 or z = 2 + J 4 - x 2 - y 2 • Therefore D is the region inside the c~c le x 2 + y 2 + (3 - 2) 2 = 4,
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