# 101 c y j d a x i b x y x i j y i j f igure 131 the

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Chapter 3 / Exercise 27
Intermediate Algebra
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101
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Chapter 3 / Exercise 27
Intermediate Algebra
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c y j d a x i b x y ( x * i j , y * i j ) F igure 13.1. The sample points and solid volume of a Riemann sum 3. If f ( x , y ) 0 on R , then RR R f ( x , y ) dA represents the volume of the solid whose upper boundary is the surface z = f ( x , y ) and whose lower boundary is R . That is, ZZ R f ( x , y ) dA = vol. ( x , y , z ) : ( x , y ) R , 0 z f ( x , y ) = vol. ( x , y , z ) : a x b , c y d , 0 z f ( x , y ) . 4. From now on, we shall use the notation | D | to denote the area of a plane region D . Let us divide both sides of (13.1) by | R | = ( b - a )( c - d ). Since Δ A = | R | / n 2 , we have Δ A / | R | = 1 / n 2 , so we get 1 | R | ZZ R f ( x , y ) dA = lim n →∞ 1 n 2 n X i = 1 n X j = 1 f ( x * i j , y * i j ) . Since 1 n 2 i j f ( x * i j , y * i j ) is the average of the sample values f ( x * i j , y * i j ), we interpret the last equa- tion in terms of the average value of f : average ( x , y ) R f ( x , y ) = 1 | R | ZZ R f ( x , y ) dA . (13.2) E xample 13.1. Evaluate the double integral of the function f ( x , y ) = ( 5 - 2 x - 3 y if 5 - 2 x - 3 y 0 , 0 if 5 - 2 x - 3 y 0 , over the rectangle R = [0 , 3] × [0 , 2]. S olution . We shall use the interpretation of the double integral as volume. By Remark 3 above, the value of the integral ZZ R f ( x , y ) dA is the volume of the solid E that lies between R and the part of the surface z = f ( x , y ) with ( x , y ) R . The graph of f ( x , y ) (before the restriction to R ) consists of the part of the plane z = 5 - 2 x - 3 y with 5 - 2 x - 3 y 0 and the part of the xy -plane z = 0 with 5 - 2 x - 3 y 0. The line with equation 5 - 2 x - 3 y = 0 splits the rectangle R in two parts: a triangle with vertices (0 , 0), ( 5 2 , 0), and (0 , 5 3 ), 102
(0 , 0) (3 , 0) (3 , 2) (0 , 2) ( 5 2 , 0) (0 , 5 3 ) F igure 13.2. The pyramid E and its base where 5 - 2 x - 3 y 0; and a pentagon with vertices ( 5 2 , 0), (3 , 0), (3 , 2), (0 , 2), and (0 , 5 3 ), where 5 - 2 x - 3 y 0 (see Figure 13.2). For points ( x , y ) lying in the pentagon, the graph z = f ( x , y ) and the xy -plane coincide, and for ( x , y ) in the triangle, the graph z = f ( x , y ) is a triangle in the plane z = 5 - 2 x - 3 y . Thus, the solid E is the pyramid with vertices O (0 , 0 , 0), A ( 5 2 , 0 , 0), B (0 , 5 3 , 0), and C (0 , 0 , 5). Its volume is 1 3 | OC | · area( 4 OAB ) = 1 3 | OC | · 1 2 | OA | | OB | = 1 3 5 · 1 2 5 2 5 3 = 125 36 . Hence, ZZ R f ( x , y ) dA = 125 36 . 13.2. Iterated integrals In practice, we want to reverse the order of things in the above example, so that we are able to compute volumes, averages, etc. by interpreting them as double integrals and then evaluating the latter using some “standard” machinery independent of our interpretation. One such method is the method of iterated integrals. An iterated integral is an expression of one of the forms Z d c Z b a f ( x , y ) dx dy or Z b a Z d c f ( x , y ) dy dx . Sometimes, we write these integrals as Z d c Z b a f ( x , y ) dxdy and Z b a Z d c f ( x , y ) dydx , respectively. Note that these are definite integrals of functions defined by means of definite inte- grals and not double integrals. Thus, we can evaluate such integrals using the familiar methods from single-variable calculus.
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