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

SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS 0 323
15. r ( u., v) = sin v i+ cos U. sin 2v j +sin u. sin 2v k. Parametric equations for the surface are x = sin v, y = cos u. sin 2v,
z = sin u. sin 2v. lfv = vo is fixed, then x = sin vo is constant, andy= {sin 2vo) cosu and z = (sin 2v 0 ) sin u. describe a
circle of radius !sin 2vol, so each-corresponding grid curve is a circle contained in the vertical plane x =sin v 0 parallel to the
yz-plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to
holding u constant, in which casey = (cosu.o) sin 2v, z =(sin uo) sin 2v ~ z =(tan uo)y, so each grid curve lies in a
plane z = ky that includes the x-axis.
17. x = cos 3 u cos 3 v, y·= sin 3 u cos 3 v, z = sin 3 v. lfv = vo is held constant then z = sin 3 v 0 is constant, so the
corresponding grid curve lies in a horizontal plane. Several of the graphs exhibit horizontal grid curves, but the curves for this
surface are neither circles nor straight lines, so graph Ill is the only possibility. (ln fact, the horizontal grid curves here are
members of the family x = acos 3 u, y = asin 3 u. and are called astroids.) The vertical grid curves we see on the surface
correspond to u = uo held constant, as then we have x = cos 3 uo cos 3 v, y = sin 3 u.o cos 3 v so the corresponding grid curve
lies in the vertical plane y = (tan 3 uo)x through the z-axis.
19. From Example 3, parametric equations for the plane through the point (0, 0, 0) that contains the vectors a = (1, -1, O) and
b = (0, 1, -1) are x = 0 + u(1) + v(O) = u., y = 0 + u.( - 1) + v(1) = v - u., z = 0 + u.(O) + v( - 1) = - v.
21. Solving the equation for x gives x 2 = 1 + y 2 + -!z 2 ~ x = J1 + y 2 + tz 2 • (We choose the positive root since we want
the part of the hyperboloid that corresponds to x ~ 0.) lfwc let y and z be the parameters, parametric equations are y = y,
z=z, x= J1+y 2 +i·z 2 •
23. Since the cone intersects the sphere in the circle x 2 + y 2 = 2, z = .../2 and we want the portion of the sphere above this, we
can parametrize the surface as x = x, y = y, z = v' 4 - x 2 - y 2 where x 2 + y 2 ~ 2.
Alternate solution: Using spherical coordinates, X = 2 sin ¢ cos e, y = 2 sin¢ sine, z = 2 cos¢ where 0 s;