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

336 0 CHAPTER 16 VECTOR CALCULUS
13. F ( x, y , z) = x J x 2 + y 2 + z 2 i + yJ x 2 + y 2 + z 2 j + z J x 2 + y 2 + z 2 k , so
div F = x. ~(x2 + y2 + z2)-112(2x) + (x2 + y2 + z2)112 + y. t(x2 + y2 + z2)-112(2y) + (x2 + y2 + z2)1/2
Then
+ z . ~(x2 + y2 + z2) - 112(2z) + (x2 + y2 + z2)1/2
= (x2 + y2 + z2)-112 [x2 + (x2 + y2 !- z2) + y2 + (x2 + y2 + z2) + z2 + (x2 + y2 + z2)]
4(x2 + y2 + z2)
= = 4Jx2+y2+z2.
Jx2 +y2 +z2
!!L
j·Tr/7:12,-11
= 4J.x 2 + y2 + z2 dV = 4.J(li . p 2 sin¢ dp dB d¢
• E 0 0 0
= f 0
1r 12 sin¢d¢J;"' d1J J;4p 3 dp = [-cos¢] ~ 12 [B]~Tr (p 4 ) ~ = (1) (27r) (1) = 21r
' 4 4 )
15. .[J s F ·dS=JJJ -/3-x 2 dV=J 1 J 1 f 2 -"' -u· v'3-x2dzdydx = 341 .J2+lllsin- 1 E - 1 - 1 Jo ao 20
(:4 a
17. For 81 we haven = - k, so F · n = F · ( -k) = - x 2 z- y 2 = -y 2 (since z = 0 on 81). So if Dis the unit disk, we get .
.f Is 1
F · dS = .ff 51
F · n dS = I J'v (-y 2 2
) dA = - f 1r It r·2 0
( sin 2 fJ) r dr dB = - {1r. Now since Sz is closed, we can use
the Divergence Theorem. Since div F = :., ( z 2 x) + tu ( iv 3 + tan z) + tz ( x 2 z + y 2 ) = z 2 + y 2 + x 2 , we use spherical
coordinates to g~t JJ 52
F · dS = I.JJ E divF dV = J 02
1r Io1r! 2 I: p 2 · p 2 sindpddO =~'IT. Finally
19. The vectors that end near P1 are longer than the vectors that start near P1, so the net flow is inward near Pt and div F(Pl) is
negative. The vectors that end near P2 are shorter than the vectors that start near P2, so the net flow is outward near P2 and
21.
div F(P2) is positiye.
5
,,,,,,,,,/
' ' ' \ 1 I f I I /
-5~-----+------45
From the graph it appears that for points above the x-axis, vectors starting near a
particular point are longer than vectors ending there, so divergence is positive.
The opposite is true at points below the x-axis, where divergence is negative.
F(x,y) = (xy, x +y 2 ) =} divF = : , (xy)+ :Y (x+y 2 ) =y+2y =Jy.
I \ ' ' '
/ ~ I I I \ \ \ '
/'/If \\\'\
-5
Thus div F > 0 for y > 0, and div F < 0 for y < 0.
. X x i+yj+ z k 8 ( X ) (x 2 +y 2 +z 2 )-3x 2 . . . .
23. Smcc - 3 = ( 2 2 2 ) 312 and - (
8 2 2 2 ) 31 ., = ( 2
.,
lxl X + y + Z X X + y + Z - X + y- + Z 2 ) 512 With Similar expressiOns
for ..!!_ ( y ) and ..!!_ ( z ) , we have
8y (x2 + y2 + z2)3/2 8z (x2 + y2 + z2)3/2
(
x ) 3(x2 + y2 + z2)- 3{x2 + y2 + z2) .
div -
1
l 3 = .. 12 · = 0, except at {0, 0, 0) where it is undefined.
x
(x2+y2+z2)"
25 . .f Is a · n dS = J I IE diva dV = 0 since div a = 0.
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