Solução_Calculo_Stewart_6e

F.
282 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
TX.10
17.5 Curl and Divergence ET 16.5
i j k
1. (a) curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z
=(−x 2 − 0) i − (−2xy − xy) j +(0− xz) k
xyz 0 −x 2 y
= −x 2 i +3xy j − xz k
(b) div F = ∇ · F = ∂
∂x (xyz)+ ∂ ∂y (0) + ∂ ∂z (−x2 y)=yz +0+0=yz
i j k
3. (a) curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z
1 x + yz xy − √ =(x − y) i − (y − 0) j +(1− 0) k
z
=(x − y) i − y j + k
(b) div F = ∇ · F = ∂
∂x (1) + ∂ ∂y (x + yz)+ ∂
xy − √
z = z − 1
∂z
2 √ z
i j k
∂/∂x ∂/∂y ∂/∂z
5. (a) curl F = ∇ × F =
(b) div F = ∇ · F = ∂
∂x
x
x2 + y 2 + z 2
y
x2 + y 2 + z 2
z
x2 + y 2 + z 2
1
=
[(−yz + yz) i − (−xz + xz) j +(−xy + xy) k] =0
(x 2 + y 2 + z 2 )
3/2
x
+ ∂ y
+ ∂
x2 + y 2 + z 2 ∂y x2 + y 2 + z 2 ∂z
z
x2 + y 2 + z 2
= x2 + y 2 + z 2 − x 2
(x 2 + y 2 + z 2 ) 3/2 + x2 + y 2 + z 2 − y 2
(x 2 + y 2 + z 2 ) 3/2 + x2 + y 2 + z 2 − z 2
(x 2 + y 2 + z 2 ) 3/2 = 2x2 +2y 2 +2z 2
(x 2 + y 2 + z 2 ) 3/2 = 2
x2 + y 2 + z 2
i j k
7. (a) curl F = ∇ × F =
∂/∂x ∂/∂y ∂/∂z
ln x ln(xy) ln(xyz)
xz
yz
y 1
=
xyz − 0 i −
xyz − 0 j +
xy − 0 k =
y , − 1 x , 1
x
(b) div F = ∇ · F = ∂
∂x (ln x)+ ∂ ∂y (ln(xy)) + ∂ ∂z (ln(xyz)) = 1 x + x xy + xy
xyz = 1 x + 1 y + 1 z
9. If the vector field is F = P i + Q j + R k, then we know R =0. In addition, the x-component of each vector of F is 0,so
P =0, hence ∂P
∂x = ∂P
∂y = ∂P
∂z = ∂R
∂x = ∂R
∂y = ∂R
∂Q
=0. Q decreases as y increases, so < 0,butQ doesn’t change
∂z ∂y
in the x-orz-directions, so ∂Q
∂x = ∂Q
∂z =0.
(a) div F = ∂P
∂x + ∂Q
∂y + ∂R
(b) curl F =
∂R
∂y − ∂Q
∂z
∂z =0+∂Q ∂y +0< 0
∂P
i +
∂z − ∂R
j +
∂x
∂Q
∂x − ∂P
∂y
k =(0− 0) i +(0− 0) j +(0− 0) k = 0