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

Section 15.1 Double Integrals over Rectangles 989
as in Figure 3, we form the subrectangles
R ij − fx i21 , x i g 3 fy j21 , y j g − hsx, yd | x i21 < x < x i , y j21 < y < y j j
each with area DA − Dx Dy.
y
d
R ij
(x i , y j )
Îy
y j
y
j-1
(x * ij , y* ij )
›
c
(x * £, y * £)
FIGURE 3
Dividing R into subrectangles
0 a ⁄ ¤
If we choose a sample point sx ij *, y ij *d in each R ij , then we can approximate the part of
S that lies above each R ij by a thin rectangular box (or “column”) with base R ij and height
f sx ij *, y ij *d as shown in Figure 4. (Compare with Figure 1.) The volume of this box is the
height of the box times the area of the base rectangle:
x i-1
Îx
x i
b
x
f sx ij *, y ij *d DA
If we follow this procedure for all the rectangles and add the volumes of the corresponding
boxes, we get an approximation to the total volume of S:
3 V < o m
i−1
o n
f sx ij *, y ij *d DA
j−1
(See Figure 5.) This double sum means that for each subrectangle we evaluate f at the
chosen point and multiply by the area of the subrectangle, and then we add the results.
z
z
a
0
c
f(x * ij,
y * ij)
d
y
0
y
x
b
x
R ij
FIGURE 4 FIGURE 5
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.