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

section 16.6 Parametric Surfaces and Their Areas 1119
z
9
D
3
x
FIGURE 16
y
Figure 16.) Using Formula 9, we have
A − y Î1 1S
Dy −xD
−z 2
1S
−yD
−z 2
dA
− yy s1 1 s2xd 2 1 s2yd 2 dA
D
− yy s1 1 4sx 2 1 y 2 d dA
D
Converting to polar coordinates, we obtain
A − y 2
0
y 3
s1 1 4r 2 r dr d − y 2
0 0
d y 3
rs1 1 4r 2 dr
− 2( 1 8) 2 3 s1 1 4r 2 d 3y2 g 0
3
− 6 (37s37 2 1) ■
0
The question remains whether our definition of surface area (6) is consistent with the
surface area formula from single-variable calculus (8.2.4).
We consider the surface S obtained by rotating the curve y − f sxd, a < x < b, about
the x-axis, where f sxd > 0 and f 9 is continuous. From Equations 3 we know that parametric
equations of S are
x − x y − f sxd cos z − f sxd sin a < x < b 0 < < 2
To compute the surface area of S we need the tangent vectors
r x − i 1 f 9sxd cos j 1 f 9sxd sin k
Thus
and so
r − 2f sxd sin j 1 f sxd cos k
Z Z
i j k
r x 3 r − 1 f9sxd cos f9sxd sin
0 2fsxd sin fsxd cos
− f sxd f 9sxd i 2 f sxd cos j 2 f sxd sin k
| r x 3 r | − sf f sxdg2 f f 9sxdg 2 1 f f sxdg 2 cos 2 1 f f sxdg 2 sin 2
because f sxd > 0. Therefore the area of S is
− sf f sxdg 2 f1 1 f f 9sxdg 2 g − f sxds1 1 f f 9sxdg 2
A − yy | r x 3 r | dA
D
− y 2
0
y b
a
f sxds1 1 f f 9sxdg 2 dx d
− 2 y b
f sxds1 1 f f 9sxdg 2 dx
a
This is precisely the formula that was used to define the area of a surface of revolution in
single-variable calculus (8.2.4).
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