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

992 Chapter 15 Multiple Integrals
x
S
FIGURE 9
z
(0, 0, 1)
(1, 0, 0) (0, 2, 0)
y
SOLUTIon It would be very difficult to evaluate this integral directly from Defini-
tion 5 but, because s1 2 x 2 > 0, we can compute the integral by interpreting it as a
volume. If z − s1 2 x 2 , then x 2 1 z 2 − 1 and z > 0, so the given double integral
represents the volume of the solid S that lies below the circular cylinder x 2 1 z 2 − 1
and above the rectangle R. (See Figure 9.) The volume of S is the area of a semicircle
with radius 1 times the length of the cylinder. Thus
The Midpoint Rule
yy s1 2 x 2 dA − 1 2 s1d2 3 4 − 2
R
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The methods that we used for approximating single integrals (the Midpoint Rule, the
Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we
consider only the Midpoint Rule for double integrals. This means that we use a double
Rie mann sum to approximate the double integral, where the sample point sx ij *, y ij *d in R ij
is chosen to be the center sx i , y j d of R ij . In other words, x i is the midpoint of fx i21 , x i g and
y j is the midpoint of fy j21 , y j g.
Midpoint Rule for Double Integrals
yy f sx, yd dA < o m
o n
f sx i , y j d DA
R
i−1 j−1
where x i is the midpoint of fx i21 , x i g and y j is the midpoint of fy j21 , y j g.
Example 3 Use the Midpoint Rule with m − n − 2 to estimate the value of the
integral yy R
sx 2 3y 2 d dA, where R − hsx, yd | 0 < x < 2, 1 < y < 2j.
y
2
3
2
1
R¡
R¡¡
R
R¡
(2, 2)
SOLUTIon In using the Midpoint Rule with m − n − 2, we evaluate f sx, yd − x 2 3y 2
at the centers of the four subrectangles shown in Figure 10. So x 1 − 1 2 , x 2 − 3 2 , y 1 − 5 4 ,
and y 2 − 7 4 . The area of each subrectangle is DA − 1 2 . Thus
yysx 2 3y 2 d dA < o 2
o 2
f sx i , y j d DA
R
i−1 j−1
− f sx 1 , y 1 d DA 1 f sx 1 , y 2 d DA 1 f sx 2 , y 1 d DA 1 f sx 2 , y 2 d DA
0
1 2
x
− f ( 1 2 , 5 4 ) DA 1 f ( 1 2 , 7 4 ) DA 1 f ( 3 2 , 5 4 ) DA 1 f ( 3 2 , 7 4 ) DA
FIGURE 10
− (2 67
16 ) 1 139
2 1 (2 16 ) 1 51
2 1 (216) 1 123
2 1 (2 16 ) 1 2
− 2 95
8 − 211.875
Number of
subrectangles
Midpoint Rule
approximation
1 211.5000
4 211.8750
16 211.9687
64 211.9922
256 211.9980
1024 211.9995
Thus we have
yy sx 2 3y 2 d dA < 211.875
R
Note In Example 5 we will see that the exact value of the double integral in Example
3 is 212. (Remember that the interpretation of a double integral as a volume is valid
only when the integrand f is a positive function. The integrand in Example 3 is not a
positive function, so its integral is not a volume. In Examples 5 and 6 we will discuss
how to interpret integrals of functions that are not always positive in terms of volumes.)
If we keep dividing each subrectangle in Figure 10 into four smaller ones with similar
shape, we get the Midpoint Rule approximations displayed in the chart in the margin.
Notice how these approximations approach the exact value of the double integral, 212.
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