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

Section 16.6 Parametric Surfaces and Their Areas 1117
P ij
Î√ r * √
(a)
S ij
Îu r * u
The part S ij of the surface S that corresponds to R ij is called a patch and has the point
P ij with position vector rsu i *, v j *d as one of its corners. Let
r u * − r u su i *, v j *d and r v * − r v su i *, v j *d
be the tangent vectors at P ij as given by Equations 5 and 4.
Figure 15(a) shows how the two edges of the patch that meet at P ij can be approximated
by vectors. These vectors, in turn, can be approximated by the vectors Du r u * and
Dv r v * because partial derivatives can be approximated by difference quotients. So
we approxi mate S ij by the parallelogram determined by the vectors Du r u * and Dv r v *.
This parallelogram is shown in Figure 15(b) and lies in the tangent plane to S at P ij . The
area of this parallelogram is
| sDu r u*d 3 sDv r*d v | − | r u* 3 r* v | Du Dv
and so an approximation to the area of S is
o m
o n
| r u* 3 r v * | Du Dv
i−1 j−1
Our intuition tells us that this approximation gets better as we increase the number of
subrectangles, and we recognize the double sum as a Riemann sum for the double integral
yy D | r u 3 r v | du dv. This motivates the following definition.
(b)
6 Definition If a smooth parametric surface S is given by the equation
FIGURE 15
Approximating a patch by a
parallelogram
rsu, vd − xsu, vd i 1 ysu, vd j 1 zsu, vd k
su, vd [ D
and S is covered just once as su, vd ranges throughout the parameter domain D,
then the surface area of S is
where
r u − −x
−u i 1 −y
−u j 1 −z
−u k
AsSd − yy | r u 3 r v | dA
D
r v − −x
−v i 1 −y
−v j 1 −z
−v k
ExamplE 10 Find the surface area of a sphere of radius a.
SOLUtion In Example 4 we found the parametric representation
x − a sin cos y − a sin sin z − a cos
where the parameter domain is
D − hs, d | 0 < < , 0 < < 2j
We first compute the cross product of the tangent vectors:
Z
Z
i j k
Z
−x −y −z
i
j
r 3 r − − − − − a cos cos a cos sin
−x −y −z 2a sin sin a sin cos
− − −
− a 2 sin 2 cos i 1 a 2 sin 2 sin j 1 a 2 sin cos k
k
2a sin
0
Z
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