Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 17.5 CURL AND DIVERGENCE ET SECTION 16.5 ¤ 283
11. If the vector field is F = P i + Q j + R k, then we know R =0. In addition, the y-component of each vector of F is 0,so
Q =0,hence ∂Q
∂x = ∂Q
∂y = ∂Q
∂z = ∂R
∂x = ∂R
∂y = ∂R
∂P
=0. P increases as y increases, so
∂z ∂y
the x-orz-directions, so ∂P
∂x = ∂P
∂z =0.
(a) div F = ∂P
∂x + ∂Q
∂y + ∂R
∂R
(b) curl F =
∂y − ∂Q
∂z
Since ∂P
∂y
∂z =0+0+0=0
∂P
i +
∂z − ∂R
∂x
∂Q
j +
∂x − ∂P
k =(0− 0) i +(0− 0) j +
∂y
> 0, −∂P k is a vector pointing in the negative z-direction.
∂y
> 0,butP doesn’t change in
0 − ∂P
∂y
k = − ∂P
∂y k
i j k 13. curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z =(6xyz 2 − 6xyz 2 ) i − (3y 2 z 2 − 3y 2 z 2 ) j +(2yz 3 − 2yz 3 ) k = 0
y 2 z 3 2xyz 3 3xy 2 z 2
and F is definedonallofR 3 with component functions which have continuous partial derivatives, so by Theorem 4,
F is conservative. Thus, there exists a function f such that F = ∇f. Thenf x(x, y, z) =y 2 z 3 implies
f(x, y, z) =xy 2 z 3 + g(y, z) and f y(x, y, z) =2xyz 3 + g y(y, z). Butf y(x, y, z) =2xyz 3 ,sog(y, z) =h(z) and
f(x, y, z) =xy 2 z 3 + h(z). Thus f z (x, y, z) =3xy 2 z 2 + h 0 (z) but f z (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.
i j k 15. curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z =(2y − 2y) i − (0 − 0) j +(2x − 2x) k = 0, F is definedonallofR 3 ,
2xy x 2 +2yz y 2
and the partial derivatives of the component functions are continuous, so F is conservative. Thus there exists a function f
such that ∇f = F. Thenf x(x, y, z) =2xy implies f(x, y, z) =x 2 y + g(y, z) and f y(x, y, z) =x 2 + g y(y, z). But
f y(x, y, z) =x 2 +2yz,sog(y, z) =y 2 z + h(z) and f(x, y, z) =x 2 y + y 2 z + h(z). Thusf z(x, y, z) =y 2 + h 0 (z) but
f z(x, y, z) =y 2 so h(z) =K and f(x, y, z) =x 2 y + y 2 z + K.
i j k
17. curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z
=(0− 0) i − (0 − 0) j +(−e −x − e −x ) k = −2e −x k 6=0,
ye −x e −x 2z
so F is not conservative.
19. No. Assume there is such a G. Thendiv(curl G) = ∂
∂x (x sin y)+ ∂ ∂y (cos y)+ ∂ (z − xy) =siny − sin y +16= 0,
∂z
which contradicts Theorem 11.
i j k
21. curl F =
∂/∂x ∂/∂y ∂/∂z
=(0− 0) i +(0− 0) j +(0− 0) k = 0. Hence F = f(x) i + g(y) j + h(z) k
f(x) g(y) h(z)
is irrotational.