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

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328 0 CHAPTER 16 VECTOR CALCULUS
16.7 Surface Integrals
1. The faces of the box in the planes x = 0 and x = 2 have surface area 24 and centers (0, 2, 3), (2, 2, 3)0 The faces in y = 0 and
y = 4 have surface area 12 and centers (1, 0, 3), (1, 4, 3), and the faces in z = 0 and z = 6 have area 8 and centers (1, 2, 0),
(1, 2, 6)0 For each face we take the point P;j to beth~ center of the face and f(x, y, z) = e -ool(x+u+z), so by Definition I,
ffs f(x, y, z) dS ~ [f(O, 2, 3)](24) + [!(2, 2, 3)](24) + [f(l, b, 3)](12)
+ [f(l, 4, 3)](12) + [f(l, 2, 0)](8) + [f(l, 2, 6)](8)
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30 We can use the xz- and yz-planes to divide H into four patches of equal size, each with surface area equal to ~ the surface
area of a sphere with radius VSO, so D.S = H4)7r( v'50) 2 = 257l'. Then (±3, ±4, 5) are sample points in the four patches,
and using a Riemann sum as in Definition l, we have
ffH f(x, y , z) dS ~ j(3, 4, 5) D.S + f(3, -4, 5) !:!.S + !( -3, 4, 5) !:!.S + f( -3, -4, 5) D.S
= (7 + 8 + 9 + 12}(257!') = 9007l' ~ 2827
So r( u, v) = ( u + v) i + ( u - v) j + (1 + 2u + v) k, 0 ~ u ~ 2, 0 ~ v ~ 1 and
0
y'3 2 + 1 2 + (- 2)2 = -1140 Then by Formula2,
r,. x r 11 = (i + j + 2k) X (i -j + k) = 3i+j - 2k ==? iru X rvl =
ff 8
(x + y + z) dS = ffv(u + v + u- v +1 + 2u + v) lru x r vl dA = f 0
1
f~(4u + v + 1)
0
-/I4dudv
= .;I4 J; [2u 2 + uv + u] ::~ dv = .;I4 J; (2v + 10} dv = .;I4 [v 2 + lOv] ~ = 11 v'l4
7o r(u,v) = (ucosv,usinov,v), 0 ~ u ~ 1, 0 ~ v ~ 7l' and
ru X rv = (cosv,sinv, O) X (-'-usinv,ucosv, 1) = (sinv, - cosv,u ) · =?
!ru X rv! = Vsin 2 v + cos 2 v + u 2 = ..)u 2 + 1. Then
ffsoY dS = ffv (usin v) lru x r vl dA = J; j~" (usinv)
0
= [ t(u 2 + 1?/ 2 ] : [- cosv]~ = t(2 3 / 2 - 1) 0 2 = H2v'2 - 1)
8z
9. z = 1 + 2x + 3y so ox = 2 and oy = 30 Then by Fonnula 4,
8z
vu 2 + 1 dv du = J; u..Ju 2 + 1 du J 0
" sin v dv
= -/14 J:j J;(x2y + 2X3?i + 3x2y2) dy dx = -/14 J; [~ x2y2 + x3y2 + x2y3]~=~ dx
= v'l4 J;(10x 2 + 4x 3 ) dx = v'l4 [ 4fx 3 + x 4 ] ~ = 171 v'f.1
110 An equation of the plane through the points (1, 0, 0), (0, -2, 0), and (0, 0, 4) is 4x- 2y + z = 4, so Sis the region in the
plane z = 4 - 4x + 2y over D = {(x, y) I 0 ~ x ~ 1,2x- 2 ~ y ~ 0}. Thus by Formula4,
ffs X dS = ffv x y'( - 4) 2 + (2) 2 1
+ 1 dA = ...J2f J; J 2 °x_ 2 x dy dx = v'2f f 0 [xy]~~~x- 2
= v'2f J 01
(- 2x 2 + 2x) dx = v'2f [-ix 3 + x 2 ]~ = v'2f (- i + 1) = 4J.
dx
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