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

318 0 CHAPTER 16 VECTOR CALCULUS
(b) curlF = ({)R - {)Q) i + ({)P - {)R) j + ({)Q - {)p) k = (0- 0) i -t (0 - O) j + (o- {)P) k =-{)P k
8y .{)z {)z 8x ax 8y 8y 8y
: fJP 0 8Pk . . . . th . d" .
S
uy vy • ·
mce -;:;- > , - .q,, IS a vector porntmg m e negative z- 1rect1on.
13. curiF = "il x F = 8f8x 8f8y
j
k
8f8z = (6xyz 2 - 6xyz 2 ) i - (~y 2 z 2 - 3y 2 z 2 )j + (2yz 3 - 2yz 3 ) k = 0
and F is defined on all oflR 3 with component functions which have continuous partial derivatives, so by Theorem 4,
F is conservativ~. Thus, there exists a· function f such that F = "il f. Then f., ( x, y , z) = y 2 z 3 implies
f(x,y,z) = xy 2 z 3 + g(y,z) and f 11 (x,y,z) = 2xyz 3 + g 11 (y,z). But f 11 (x,y,z) = 2xyz 3 ,sog(y,z) = h(z) and
f(x, y,z) = xy 2 z 3 + h(z). Thus fz(x, y, z) = 3xy 2 z 2 + h'(z) but f:(x, y, z) = 3xy 2 z 2 so· h(z) = K, a constant.
Hence a potential function for F is f (x, y, z) = xy 2 z 3 + K.
15. curl F = "il X F = 8f8x 8f8y 8f8z
j
3xy 2 z 2 2x 2 yz 3 3x 2 y 2 z 2
= (6x 2 yz 2 - 6x 2 yz 2 ) i - (6xy 2 z 2 - 6xy 2 z) j .+ (4xyz 3 - 6xyz 2 ) k
= 6xy 2 z(I- z) j + 2xyz 2 (2z- 3) k =1- 0
so F is not conservative.
17. cur!F = "il x F = 8f8x 8f8y 8f8z
j
k
k
F is defined on all oflll 3 , and the partial derivatives of the component functi~ns are continuous,'so F is conservative. Thus
there exists a function f such that "il f =F. Then f,(x, y, z) = e 11 = implies f(x, y, z) = xe 11 = + g(y, z) =>
fu(x, y, z) = xzev= + g 11 (y, z). But / 11
{x, y, z) = xze 11 =, so g(y, z) = h(z) and f(x, y, z) = xe 11 z + h(z).
Thus f:(x, y, z) = xye 11 z + h'(z) but f:(x, y, z) = xye 11 = so h(z) =Kanda potential function for F is
f(x, y, z) = xe 11 = + K .
. 19. No. Assume there is such a G. Then div(curl G ) ='! (x sin y) + ~ (cosy) + ~ (z - xy) =sin y- siny + 1 f 0,
uX uy uz
which contradicts Theorem 11 .
j
k
21 . curlF = 8f8x 8f8y 8f8z = {0- 0) i + {0- O)j + {0 - 0) k = 0. Hence F = f(x) i + g(y) j + h(z) k
is irrotational.
f(x) g(y) h(z)
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