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

988 Chapter 15 Multiple Integrals
In much the same way that our attempt to solve the area problem led to the definition of
a definite integral, we now seek to find the volume of a solid and in the process we arrive
at the definition of a double integral.
Review of the Definite Integral
First let’s recall the basic facts concerning definite integrals of functions of a single variable.
If f sxd is defined for a < x < b, we start by dividing the interval fa, bg into n subintervals
fx i21 , x i g of equal width Dx − sb 2 adyn and we choose sample points x* i in
these subintervals. Then we form the Riemann sum
1 o n
i−1
f sx i *d Dx
and take the limit of such sums as n l ` to obtain the definite integral of f from a to b:
2 y b
a
f sxd dx − lim
n l ` on f sx*d i Dx
i−1
In the special case where f sxd > 0, the Riemann sum can be interpreted as the sum of
the areas of the approximating rectangles in Figure 1, and y b a
f sxd dx represents the area
under the curve y − f sxd from a to b.
y
Îx
f(x i *)
0
a
⁄ ¤ ‹ x i-1 x i x n-1
b
x
FIGURE 1
x¡* x* x£* x i
*
x n
*
z
z=f(x, y )
Volumes and Double Integrals
In a similar manner we consider a function f of two variables defined on a closed rectangle
R − fa, bg 3 fc, dg − hsx, yd [ R 2 | a < x < b, c < y < dj
0
a
b
x
c
R
d
y
and we first suppose that f sx, yd > 0. The graph of f is a surface with equation z − f sx, yd.
Let S be the solid that lies above R and under the graph of f , that is,
S − hsx, y, zd [ R 3 | 0 < z < f sx, yd, sx, yd [ Rj
FIGURE 2
(See Figure 2.) Our goal is to find the volume of S.
The first step is to divide the rectangle R into subrectangles. We accomplish this by
dividing the interval fa, bg into m subintervals fx i21 , x i g of equal width Dx − sb 2 adym
and dividing fc, dg into n subintervals fy j21 , y j g of equal width Dy − sd 2 cdyn. By
drawing lines parallel to the coordinate axes through the endpoints of these subintervals,
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