Explanation Given: The surface f ( x , y ) = x y and R is the triangle bounded by y = x , y = 2 − x and y = 0 . Formula used: The iterated integral is evaluated as ∬ R f ( x , y ) d A = ∫ a b ( ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y ) d x , where f is a continuous function, g 1 and g 2 are continuous defined as g 1 ( x ) ≤ g 2 ( x ) on the interval [ a , b ] and R = { ( x , y ) | g 1 ( x ) ≤ y ≤ g 2 ( x ) , a ≤ x ≤ b } . The average value of f ( x , y ) over the region on R is ∬ R f ( x , y ) d A ∬ R d A . Calculation: Use online graphing calculator to sketch the region bounded by the lines y = x , y = 2 − x and y = 0 . From Figure 1 observe that the shaded region is the required region. The triangle has been divided into two short triangles. Therefore, the limit has to be divided for the two triangles. Thus, the average value of the surface f ( x , y ) = x y over the region is given by ∫ 0 1 ∫ 0 x ( x y d y ) d x ∫ 0 1 ∫ 0 x ( d y ) d x + ∫ 1 2 ∫ 0 2 − x ( x y d y ) d x ∫ 1 2 ∫ 0 2 − x ( d y ) d x . Consider the triangle with vertices ( 0 , 0 ) , ( 1 , 1 ) , ( 1 , 0 ) , with limits 0 ≤ x ≤ 1 , 0 ≤ y ≤ x . Substitute these values in the ∬ R d A formula used as follows. ∬ R d A = ∫ 0 1 ∫ 0 x ( d y ) d x . The first integral ∫ 0 x ( d y ) can be simplified as follows. ∫ 0 x ( d y ) = [ y ] 0 x = [ ( x ) − 0 ] = x Substitute the value of x in the second integral as follows. ∫ 0 1 [ x ] d x = [ x 2 2 ] 0 1 = [ ( 1 ) 2 2 − 0 ] = 1 2 The value of ∫ 0 1 ∫ 0 x ( d y ) d x is 1 2 . Substitute the limit values in the ∬ R f ( x , y ) d A formula used as follows. ∬ R f ( x , y ) d A = ∫ 0 1 ∫ 0 x ( x y d y ) d x . The first integral ∫ 0 x ( x y d y ) can be simplified as follows. ∫ 0 x ( x y d y ) = [ x y 2 2 ] 0 x = x ( x ) 2 2 = x 3 2 Substitute the value of x 3 2 in the second integral as follows. ∫ 0 1 [ x 3 2 ] d x = 1 2 [ x 4 4 ] 0 1 = 1 2 [ ( 1 ) 4 4 − 0 ] = 1 8 The value of ∫ 0 1 ∫ 0 x ( x y d y ) d x is 1 2 . Substitute the values ∫ 0 1 ∫ 0 x ( x y d y ) d x = 1 8 and ∫ 0 1 ∫ 0 x ( d y ) d x = 1 2 as follows. ∫ 0 1 ∫ 0 x ( x y d y ) d x ∫ 0 1 ∫ 0 x ( d y ) d x = [ 1 8 ] [ 1 2 ] = 1 8 ( 2 1 ) = 1 4 The average value of the first triangle with vertices ( 0 , 0 ) , ( 1 , 1 ) , ( 1 , 0 ) is 1 4

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 8.8, Problem 19E

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To find: The average value of the function f(x,y) over the plane region R.

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Chapter 8.8, Problem 18E

Chapter 8.8, Problem 20E

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Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

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The average value of the function f ( x , y ) over the plane region R .

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Chapter 8 Solutions