James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 16.6 Parametric Surfaces and Their Areas 1115
For some purposes the parametric
representations in Solutions 1 and 2 are
equally good, but Solution 2 might be
preferable in certain situations. If we
are interested only in the part of the
cone that lies below the plane z − 1, for
instance, all we have to do in Solution 2
is change the parameter domain to
0 < r < 1 2 0 < < 2
z
z − 2sx 2 1 y 2 − 2r. So a vector equation for the cone is
where r > 0 and 0 < < 2.
Surfaces of Revolution
rsr, d − r cos i 1 r sin j 1 2r k
Surfaces of revolution can be represented parametrically and thus graphed using a computer.
For instance, let’s consider the surface S obtained by rotating the curve y − f sxd,
a < x < b, about the x-axis, where f sxd > 0. Let be the angle of rotation as shown
in Figure 10. If sx, y, zd is a point on S, then
■
3 x − x y − f sxd cos z − f sxd sin
0
y=ƒ
y
Therefore we take x and as parameters and regard Equations 3 as parametric equations
of S. The parameter domain is given by a < x < b, 0 < < 2.
x
ƒ
x
¨
ƒ
z
(x, y , z)
ExamplE 8 Find parametric equations for the surface generated by rotating the curve
y − sin x, 0 < x < 2, about the x-axis. Use these equations to graph the surface of
revolution.
SOLUtion From Equations 3, the parametric equations are
FIGURE 10
z
y
FIGURE 11
x
x − x y − sin x cos z − sin x sin
and the parameter domain is 0 < x < 2, 0 < < 2. Using a computer to plot these
equations and and then rotating the image, we obtain the graph in Figure 11.
We can adapt Equations 3 to represent a surface obtained through revolution about the
y- or z-axis (see Exercise 30).
Tangent Planes
We now find the tangent plane to a parametric surface S traced out by a vector function
rsu, vd − xsu, vd i 1 ysu, vd j 1 zsu, vd k
at a point P 0 with position vector rsu 0 , v 0 d. If we keep u constant by putting u − u 0 , then
rsu 0 , vd becomes a vector function of the single parameter v and defines a grid curve C 1
lying on S. (See Figure 12.) The tangent vector to C 1 at P 0 is obtained by taking the partial
derivative of r with respect to v:
4 r v − −x
−v su 0, v 0 d i 1 −y
−v su 0, v 0 d j 1 −z
−v su 0, v 0 d k
■
√
z
P¸
(u ¸, √ ¸)
√=√¸
D
u=u ¸
r
r √
C¡
r u
0 u
0
C
FIGURE 12
x
y
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.