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

Section 16.7 Surface Integrals 1125
Therefore, in this case, Formula 2 becomes
4 y
S
y fsx, y, zd dS − y
D
y f sx, y, tsx, yddÎS −z
2
−xD 1S
−yD
−z 2
1 1 dA
Similar formulas apply when it is more convenient to project S onto the yz-plane or
xz-plane. For instance, if S is a surface with equation y − hsx, zd and D is its projection
onto the xz-plane, then
y f sx, y, zd dS − yy f sx, hsx, zd, zdÎS
−xD
−y 2
1S
−zD
−y 2
1 1 dA
y
S
D
z
Example 2 Evaluate yy S
y dS, where S is the surface z − x 1 y 2 , 0 < x < 1,
0 < y < 2. (See Figure 2.)
FIGURE 2
x
y
SOLUTION Since
Formula 4 gives
y
S
y y dS − y
−z
−x − 1 and −z
−y − 2y
D
y yÎ1 1S −z
2
−xD 1S
−yD
−z 2
dA
− y 1
0 y2 0
ys1 1 1 1 4y 2 dy dx
− y 1
dx s2 y 2
ys1 1 2y 2 dy
0
0
− s2 ( 1 4) 2 3 s1 1 2y 2 d 3y2 g 0
2
−
13s2
3
■
z
y
0
S
FIGURE 3
S£ (z=1+x )
S¡ (≈+¥=1)
x
If S is a piecewise-smooth surface, that is, a finite union of smooth surfaces S 1 , S 2 , . . . ,
S n that intersect only along their boundaries, then the surface integral of f over S is
defined by
yy f sx, y, zd dS − yy f sx, y, zd dS 1 ∙ ∙ ∙ 1 yy f sx, y, zd dS
S
S 1 S n
ExamplE 3 Evaluate yy S
z dS, where S is the surface whose sides S 1 are given by the
cylinder x 2 1 y 2 − 1, whose bottom S 2 is the disk x 2 1 y 2 < 1 in the plane z − 0, and
whose top S 3 is the part of the plane z − 1 1 x that lies above S 2 .
SOLUTION The surface S is shown in Figure 3. (We have changed the usual position
of the axes to get a better look at S.) For S 1 we use and z as parameters (see Example
16.6.5) and write its parametric equations as
where
x − cos y − sin z − z
0 < < 2 and 0 < z < 1 1 x − 1 1 cos
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