Solução_Calculo_Stewart_6e

F.
286 ¤ CHAPTER 17 VECTOR CALCULUS ET CHAPTER 16
i j k
(b) From (a), v = w × r =
0 0 ω
=(0· z − ωy) i +(ωx − 0 · z) j +(0· y − x · 0) k = −ωy i + ωx j
x y z
i j k
(c) curl v = ∇ × v =
∂/∂x ∂/∂y ∂/∂z
−ωy ωx 0
∂
=
∂y (0) − ∂
∂z (ωx) i +
=[ω − (−ω)] k =2ω k =2w
TX.10
∂
∂z (−ωy) − ∂
∂x (0)
j +
∂
∂x (ωx) − ∂ ∂y (−ωy)
k
39. For any continuous function f on R 3 ,defineavectorfield G(x, y, z) =hg(x, y, z), 0, 0i where g(x, y, z) = x
f (t, y, z) dt.
0
Then div G = ∂
∂x (g(x, y, z)) + ∂ ∂y (0) + ∂ ∂z (0) = ∂ x
f(t, y, z) dt = f(x, y, z) by the Fundamental Theorem of
∂x
0
Calculus. Thus every continuous function f on R 3 is the divergence of some vector field.
17.6 Parametric Surfaces and Their Areas ET 16.6
1. P (7, 10, 4) lies on the parametric surface r(u, v) =h2u +3v, 1+5u − v, 2+u + vi if and only if there are values for u
and v where 2u +3v =7, 1+5u − v =10,and2+u + v =4.Butsolvingthefirst two equations simultaneously gives
u =2, v =1and these values do not satisfy the third equation, so P does not lie on the surface.
Q(5, 22, 5) lies on the surface if 2u +3v =5, 1+5u − v =22,and2+u + v =5for some values of u and v. Solving the
first two equations simultaneously gives u =4, v = −1 and these values satisfy the third equation, so Q lies on the surface.
3. r(u, v) =(u + v) i +(3− v) j +(1+4u +5v) k = h0, 3, 1i + u h1, 0, 4i + v h1, −1, 5i. From Example 3, we recognize
this as a vector equation of a plane through the point (0, 3, 1) and containing vectors a = h1, 0, 4i and b = h1, −1, 5i. Ifwe
i j k
wish to find a more conventional equation for the plane, a normal vector to the plane is a × b =
1 0 4
=4i − j − k
1 −1 5
andanequationoftheplaneis4(x − 0) − (y − 3) − (z − 1) = 0 or 4x − y − z = −4.
5. r(s, t) = s, t, t 2 − s 2 , so the corresponding parametric equations for the surface are x = s, y = t, z = t 2 − s 2 .Forany
point (x, y, z) onthesurface,wehavez = y 2 − x 2 . With no restrictions on the parameters, the surface is z = y 2 − x 2 ,which
we recognize as a hyperbolic paraboloid.