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MULTIPLE INTEGRALS
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[CHAP. 9
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where the limit is taken so that the number n of subdivisions increases without limit and such that the largest linear dimension of each Ak approaches zero. See Fig. 9-2(a). If this limit exists, it is denoted by F x; y dA
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and is called the double integral of F x; y over the region r. It can be proved that the limit does exist if F x; y is continuous (or sectionally continuous) in r. The double integral has a great variety of interpretations with any individual one dependent on the form of the integrand. For example, if F x; y  x; y represents the variable density of a at iron plate then the double integral, A  dA, of this function over a same shaped plane region, A, is the mass of the plate. In Fig. 9-2(b) we assume that F x; y is a height function (established by a portion of a surface z F x; y for a cylindrically shaped object. In this case the double integral represents a volume.
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Fig. 9-2
ITERATED INTEGRALS If r is such that any lines parallel to the y-axis meet the boundary of r in at most two points (as is true in Fig. 9-1), then we can write the equations of the curves ACB and ADB bounding r as y f1 x and y f2 x , respectively, where f1 x and f2 x are single-valued and continuous in a @ x @ b. In this case we can evaluate the double integral (3) by choosing the regions rk as rectangles formed by constructing a grid of lines parallel to the x- and y-axes and Ak as the corresponding areas. Then (3) can be written
F x; y dx dy
f2 x F x; y dy dx ' F x; y dy dx
x a y f1 x
b & f2 x
x a y f1 x
CHAP. 9]
MULTIPLE INTEGRALS
where the integral in braces is to be evaluated rst (keeping x constant) and nally integrating with respect to x from a to b. The result (4) indicates how a double integral can be evaluated by expressing it in terms of two single integrals called iterated integrals. The process of iterated integration is visually illustrated in Fig. 9-3a,b and further illustrated as follows.
Fig. 9-3
is f2 x 0 f1 x 0 , i.e., y2 y1 . Thus, the area of the section is A
The general idea, as demonstrated with respect to a given three-space region, is to establish a plane section, integrate to determine its area, and then add up all the plane sections through an integration with respect to the remaining variable. For example, choose a value of x (say, x x 0 . The intersection of the plane x x 0 with the solid establishes the plane section. In it z F x 0 ; y is the height function, and if y f1 x and y f2 x (for all z) are the bounding cylindrical surfaces of the solid, then the width
F x 0 ; y dy. Now establish slabs
Aj xj , where for each interval xj xj xj 1 , there is an intermediate value xj0 . Then sum these to get an approximation to the target volume. Adding the slabs and taking the limit yields V lim
n X j 1
Aj xj
b  y2
a y1
 F x; y dy dx
In some cases the order of integration is dictated by the geometry. For example, if r is such that any lines parallel to the x-axis meet the boundary of r in at most two points (as in Fig. 9-1), then the equations of curves CAD and CBD can be written x g1 y and x g2 y respectively and we nd similarly F x; y dx dy
g2 y F x; y dx dy ' F x; y dx dy
y c x g1 y
d & g2 y
y c x g1 y
If the double integral exists, (4) and (5) yield the same value. (See, however, Problem 9.21.) In writing a double integral, either of the forms (4) or (5), whichever is appropriate, may be used. We call one form an interchange of the order of integration with respect to the other form.
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