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

SECTION 16.7 SURFACE INTEGRALS 0 331
On 8 1 : The surface is z = J1 - y 2 for 0 S x S 2, - 1 S y S 1 with upward orientation, so
f F . dS = 1 ~ !
1
[-x 2 (0) - y 2 ( - h) + z 2 2 1
] dy dx = 1 / ( h + 1 - y 2 ) dy dx
! "
ls1 o -1 1 -y 2 . · o - 1 1-y 2 .
On 8 2 : The surface is z = 0 with downward orientation, so
On 8 3 : The surface is x = 2 for - 1 S y S 1, O.S z S yf1- y 2 , oriented in the positive x-direction. Regarding y and z as
parameters, we have r 11 x r .., = i and
On 8 4 : The surface is x = 0 for -1 :::; y :::; 1, 0 $ z $ yf1 - y 2 , oriented in the negative x-direction. Regarding y and z as
parameters, we use - (r 11
x r z) = -i and
ff 54
F · dS = f~ 1 fo~ x 2 dzdy = f~ 1 fo~ (0) dz dy = 0
Thus ffs F · dS = 1 + 0 + 211" + 0 = 211" + ~·
33. z = xeY => 8zj8x = e 11 , 8zj8y = xe 11 , so by Formula 4, a CAS gives
JJ 5
(x 2 +y 2 +z 2 )d8 = J 01
f 1 0
(x 2 +y 2 +x 2 e 211 )Je 2 Y +x 2 e2v + 1dxdy ~ 4.5822.
35. We use Formula 4 with z = 3- 2x 2 - y 2 => 8zj8x = ·- 4x, 8zj8y = -2y. The boundaries of the region
3- 2x 2 - y 2 ~ 0 are -/i $ X s ji and -J3- 2x 2 $ y $ J3- 2.7: 2 • so we use a CAS (with precision reduced to
seven or fewer digits; otherwise the calculation may take a long time) to calculate
37. If 8 is given byy = h(x,z), then 8 is also the level surface f(x,y,z) = y - h(x,z) = 0.
Vf(x,y,z) - h., i+j-h: k d . h · 1 h · h 1ft N ·
n = I~/( )I = , an - n IS t e umt norma t at pomts tot e e . ow we proceed as m the
v x,y,z y'h~+l+h~
derivation of (1 0), using Formula 4 to evaluate
where D is the projection of 8 onto the xz-plane. Therefore I Is F · dS = I i ( P ~~ - Q + R ~~) dA.
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