1 Administrative Announcements -Midterm on Tuesday, 3/18, 6:45-8PM -Conflict, Wednesday, 3/19, 6:45-8AM -No class on 3/19, 3/21 -Homework assignment due on 3/17Copyright © Cengage Learning. All rights reserved. 15 Multiple Integrals3 3 Copyright © Cengage Learning. All rights reserved. 15.2 Iterated Integrals4 4 Iterated Integrals Suppose that f is a function of two variables that is integrable on the rectangle R = [a, b] × [c, d ]. We use the notation to mean that x is held fixed and f (x, y) is integrated with respect to y from y = c to y = d. This procedure is called partial integration with respect to y. (Notice its similarity to partial differentiation.) Now is a number that depends on the value of x, so it defines a function of x:5 5 Iterated Integrals If we now integrate the function A with respect to x from x = a to x = b, we get The integral on the right side of Equation 1 is called an iterated integral. Usually the brackets are omitted. Thus means that we first integrate with respect to y from c to d and then with respect to x from a to b.6 6 Iterated Integrals Similarly, the iterated integral means that we first integrate with respect to x (holding y fixed) from x = a to x = b and then we integrate the resulting function of y with respect to y from y = c to y = d. Notice that in both Equations 2 and 3 we work from the inside out.7 7 Example 1 Evaluate the iterated integrals. (a) (b) Solution: (a) Regarding x as a constant, we obtain8 8 Example 1 – Solution Thus the function A in the preceding discussion is given by in this example. We now integrate this function of x from 0 to 3: cont’d9 9 Example 1 – Solution (b) Here we first integrate with respect to x: cont’d10 10 Iterated Integrals Notice that in Example 1 we obtained the same answer whether we integrated with respect to y or x first. In general, it turns out (see Theorem 4) that the two iterated integrals in Equations 2 and 3 are always equal; that is, the order of integration does not matter. (This is similar to Clairaut’s Theorem on the equality of the mixed partial derivatives.)11 11 Iterated Integrals The following theorem gives a practical method for evaluating a double integral by expressing it as an iterated integral (in either order).12 12 Iterated Integrals In the special case where f (x, y) can be factored as the product of a function of x only and a function of y only, the double integral of f can be written in a particularly simple form. To be specific, suppose that f (x, y) = g (x)h (y) and R = [a, b] × [c, d]. Then Fubini’s Theorem gives13 13 Iterated Integrals In the inner integral, y is a constant, so h(y) is a constant and we can write since is a constant. Therefore, in this case, the double integral of f can be written as the product of two single integrals:Copyright © Cengage Learning. All rights reserved. 15.3 Double Integrals over General Regions15 15 Double Integrals over General Regions Learning objectives: § evaluate double integrals over regions of type I and/or type II as iterated integrals § evaluate double integrals over regions that are the union of regions of type I and/or type II § find the region of integration given by the intersection of (two or more) plane curves § find the area of plane regions § find the volume of solids above a plane region and below the graph of a function of two variables § reverse the order of integration of iterated integrals16 16 Double Integrals over General Regions For single integrals, the region over which we integrate is always an interval. But for double integrals, we want to be able to integrate a function f not just over rectangles but also over regions D of more general shape, such as the one illustrated in Figure 1. Figure 117 17 Double Integrals over General Regions A plane region D is said to be of type I if it lies between the graphs of two continuous functions of x, that is, D = {(x, y) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)} where g1 and g2 are continuous on [a, b]. Some examples of type I regions are shown in Figure 5. Some type I regions Figure 518 18 Double Integrals over General Regions The following formula enables us to evaluate the double integral over a type I region as an iterated integral. The integral on the right side of is an iterated integral, except that in the inner integral we regard x as being constant not only in f (x, y) but also in the limits of integration, g1 (x) and g2 (x).19 19 Double Integrals over General Regions We also consider plane regions of type II, which can be expressed as D = {(x, y) | c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y)} where h1 and h2 are continuous. Two such regions are illustrated in Figure 7. Some type II regions Figure 720 20 Double Integrals over General Regions Using the same methods that were used in establishing , we can show that21 21 Properties of Double Integrals22 22 Properties of Double Integrals We assume that all of the following integrals exist. The first three properties of double integrals over a region D follow immediately from Definition 2. If f (x, y) ≥ g (x, y) for all (x, y) in D, then23 23 Properties of Double Integrals The next property of double integrals is similar to the property of single integrals given by the equation If D = D1 U D2, where D1 and D2 don’t overlap except perhaps on their boundaries (see Figure 17), then Figure 1724 24 Properties of Double Integrals Property 9 can be used to evaluate double integrals over regions D that are neither type I nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illustrates this procedure. Figure 18(a) D is neither type I nor type II. D = D1 ∪ D2, D1 is type I, D2 is type II. Figure 18(b)25 25 Properties of Double Integrals The next property of integrals says that if we integrate the constant function f (x, y) = 1 over a region D, we get the area of D:26 26 Properties of Double Integrals Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose height is 1 has volume A(D) ! 1 = A(D), but we know that we can also write its volume as ∫∫D 1 dA. Cylinder with base D and height 1 Figure 1927 27 Properties of Double Integrals Finally, we can combine Properties 7, 8, and 10 to prove the following
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