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

8 8 8
(b)clivF = 'V ·iF=- (x+yz) + - (y+xz)+ - (z + xy)=1 + 1+1 =3
0 X 0 y 0 Z
j
k
3. (a) curlF = \1 x F = 8/8x 8/8y 8/8z = (ze"- 0) i- (yze"' - xye•)j + (0 - xe") k
xyez 0 yze"'
= ze"' i + (xye" - yze"' ) j - xez k
8 8 8 .
(b) clivF = 'V · F =ox (xye") + oy (0) + oz (yze'") = ye• + 0 + ye"' = y(e• + e"' )
SECTION 16.5 CURL AND DIVERGENCE D 317
5. (a) curlF = \1 x F =
j
8/8x 8/8y 8/ 8z
X y z
.jx2 + y2 + z2 .jx2 + y2 + z2
k
= ( 2
?l
X +y- + Z 2 ) 312
[(- yz+ yz) i -(-xz +xz)j + (- xy+xy) k ]=O
(b) div F = \1. F = ..E._ ( x ) + ~ ( Y ) +..E._ ( Z )
ox .j x2 + y2 + z 2 8y .j x2 + y2 + z2 az .j x2 + y2 + z2
x2 + y2 + z2 _ x2 x2 + y2 + z2 _ y2 x2 + y2 + z2 _ z 2 2x2 + 2 y2 + 2 z2 2
= (x2 + y2 + z2)3/ 2 + (x2 + y2 + z2)3/ 2 + (x2 + y2 + z2)3/ 2 = (x2 + y2 + z 2)3/ 2 = -..jr-x=;2;=+=y:;:= 2 =+=z=;;: 2
j
k
7. (a) curlF = \1 x F = 8/ ax 8/8y 8/ 8z = {0 - eV cosz) i - (e"' cosx - O)j + (0- e'" cosy) k
e"' sin y eY sin z
ez sin x
= (-e 11 cosz, - e" cosx, -e"' cosy)
(b) cliv F = \1 · F = ! ( e"' sin y) + ~ ( eV sin z) + ! ( e" sin x) = e'" sin y + eY sin z + e" sin x
9. If the vector field is F = .Pi+ Q j + Rk, then we know R = 0. ln addition, the x-component of each vector ofF is 0, so
8P 8P aP 8R 8R aR . . 8Q ·
P = 0, hence - 0
= -a = -a = - 0
= - 0
= -a = 0. Q decreases as y mcreases, so - 0
< 0, but Q doesn't change
X y Z X y Z y
. h d' . . oQ oQ 0
In t eX- Or Z · 1rectJons, SO ax = az = .
. 8P 8Q 8R 8Q
(a)cliv F = - +-+-a =0+ - + O< O
0 X 8 y Z 0 y
(b) curlF =(. oR - aQ) i + (oP- oR) j + (oQ - ap) k = (0 -O) i+ (0 - O)j + (0 - O)k = 0
8y 8z az OX ax ay
11. If the vector field is F = Pi + Q j + R k,' then we know R = 0. ln addition, the y-component of each vector ofF is 0, so
8Q 8Q 8Q 8R aR 8R . . 8P .
Q = 0, hence ox = 8y = az = ox = oy = az = 0. P mcreases as y mcreases, so oy > 0, but P doesn't change m
. th d' . ap ap O
e X- or Z - lreCtiOns, SO ax = az = .
. ap aQ aR
(a) d1v F = - + - + - = 0 + 0 + 0 = 0
ax 8y 8z
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