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

-0 PROBLEMS PLUS
1. Let 81 be the portion ofD(8) between 8(a) and 8, and let 881 be its boundary. Also let 8L be the lateral surface of 81 [that
is, the surface of 8 1 except 8 and 8(a)]. Applying the Divergence Theorem we have/" {
But
r ·3n
d8 = { { { \l. r 3
dV.
J 8S1 r J J J s 1 r
"il · :3 = \ :x 1 :y'!) · \ (x2 + y2x+ z2)3/2' (x2, + y2y + z2)3/2' (x2 + y./+ z2//2)
(x2 + 1/ + z2 - 3x2) + (x2 + y2 + z2- 3y2) + (~2 + y2 + z2 - 3z2)
- - o
- (x2 + y2 + z2)5/2 -
=> j" { r ' 3
° d8 = !"{ 0 dV = 0. On 'the other hand, notice that for the surfaces of 881 other than S(a) and 8,
l os , r J l s,
r · n = O =>
{ r ·3n d8 =>
l os, r Js r Js(a) r JsL r i s r Js(a) r
0 = !"{ r ·3n dS = !"{ r ·3n dS + !"{ r ·an d8 + !"{ r ·sn d8 = !"{ r ·3n d8 + !"
!Is !"1 - r · . - n r · n r r
3
d8 = - -
3 - d8. Notice that on S(a), r =a => n = - - =-- and r · r = r 2 = a 2 , so
s 1 S(a) r r a .
that - / " r r ·3n d8 = !" r r ·4r d8 = !" r a: d8 = ~ !" r dS = area of2S (a) = ID(8) 1.
J s(a) r . Js(a) a J s(a) a a Js(a) a
! "{
Therefore ID(S)I = Js ~ r · n dS.
3. The given line i ntegra l ~ .{ 0 (bz - CIJ) dx +(ex- az ) dy + (ay - bx) dz can be expressed as fc F · dr if we define the vector
field F by F (x, y, z) =Pi + Q j + R k = ~(bz- CIJ) i + Hex- az) j + ~(ay- bx) k. Then define 8 to be the planar
interior ofC, so Sis an oriented, smooth surface. Stokes' Theorem says .fc F · dr = .ffs curlF · dS = JJ~ curl F . n d8.
Now
curl F = ( ~: -
~~) i + (a:; - ~=) j + ( ~~ - ~:) k
= (~a+ ~a) i + (~b+ tb) j + (~c+ tc) k = a i +bj + ck = n
so curlF · n = n · n = lnl 2 = 1, hence Jfs curl F · n dS = Jfs dS w hich is simply the su rface area of S. Thus,
J~ F · dr = ~ j~(bz- cy) dx +(ex- az) dy + (ay .- bx) dz is the plane area.enclosed by C.
5. (F ·\l) G= (p~:.rc+Ql:y+R l !)cP2 i +Q2j+R2 k)
= (p 1
8P2 + Q 1
8P2 + R 1
8P2) i + (PI 8Q2 + Q 1
8Q2 + R 1
8Q2) j
ax By 8z OX 8y az
= (F · \JP2) i + (F · \JQ2) j + (F · \JR2) k.
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