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

SECTION 16.6 PARAMETRIC SURFACES AND THE;IR AREAS 0 321
j
k
(b) From(a), v=w x r= 0 0 w =( O ·z -wy)i + (wx~ O ·z) j+ ( O · y -x· O ) k=-wyi+ wx j
X y Z
(c) cur] v = "il X v = ajax a jay ajaz
j
k
-wy wx 0
= [~ (0) - ~ (wx)] i + [~ (- wy) - ~ (o)] j + [~ (wx)- ~ (-wy)] k
ay az az ax ax ay
= [w - (- w)] k = 2wk = 2w
39. For any continuous function f on iR 3 , define a vector field G ( x , y, z) = (g(x, y , z) , 0, 0) where g( x, y, z ) = f ox f ( t, y, z) dt.
a a a a x
Then div G = -a (g(x ,.y, z)) +- a (0) +- a (0) = ~ f 0
f (t , y , z) dt = f (x, y, z) by the Fundamental Theorem of
X y Z v X .
Calculus. Thus every continuous function f on R 3 is the divergence of some vector field.
16.6 Parametric Surfaces and Their Areas
1. P (7, 10, 4) lies on the parametric surface r ( u, v) = (2u + 3v , 1 + 5u - v , 2 + u + v) if and only if there are values for u
and v where 2u + 3v = 7, 1 + 5u - v = 10, and 2 + u + v = 4. But solving the first two equations simultaneously gives
u = 2, v = 1 and t~ese values do not satisfy the third equation, so P does not He on the surface.
Q(5, 22, 5) lies on the surface if2u + 3v = 5, 1 + 5u- v = 22, and 2 + u + v = 5 for some values ofu 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 = (0,3, 1} + u (1,0, 4) + v(1, -1, 5). From Example 3, we recognize
this as a vector equation of a pl.ane through the point (0, 3, 1) and containing vectors a = {1, 0, 4) and b = (1, - 1, 5). If we
j k
wish to find a more conventional equation for the plane, a normal vector to the plane is ~ X b = 1 0 4 =4i - j-k
1 - 1 5
and an equation of the plane is 4(x- 0)- (y- 3) :._ (z- 1) = 0 or 4x- y- z = - 4.
5. r(s, t ) = (s, t , e - s 2 ) , so the corresponding parametric equations for the surface are X = s, y = t, z = t 2 - s 2 . For any
point (x, y , z) on the surface, we have z = 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.
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