Explanation Given: The region is R = { ( x , y ) : x + y ≤ a , x ≥ 0 , y ≥ 0 } where a is positive constant. Calculation: Obtain the area of region R. From the given region observed that the limit of x varies from 0 to a and the limit of y varies from 0 to a − x . The integral for area of the region is A = ∫ 0 a ∫ 0 a − x d y d x . Integrate the above integral with respect to y and apply the limit. A = ∫ 0 a [ y ] 0 a − x d x = ∫ 0 a [ ( a − x ) − ( 0 ) ] d x = ∫ 0 a ( a − x ) d x Integrate the above integral with respect to x and apply the limit. A = [ a x − x 2 2 ] 0 a = [ ( a ( a ) − ( a ) 2 2 ) − ( a ( 0 ) − ( 0 ) 2 2 ) ] = [ a 2 − a 2 2 − 0 + 0 ] = a 2 2 Thus, the area of the region R is a 2 2 . By using the given formula f ¯ = 1 area of R ∬ R f ( x , y ) d A obtain the average value for the region R . The corresponding integral to obtain the average value is f ¯ = 1 ( a 2 2 ) ∫ 0 a ∫ 0 a − x ( a − x − y ) d y d x . Integrate the above integral with respect to y and apply the limit

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

2nd Edition

ISBN: 9780321954329

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 13.2, Problem 89E

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To find: The average value of a function z=a−x−y over the region R.

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Chapter 13.2, Problem 88E

Chapter 13.2, Problem 90E

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Chapter 13 Solutions

Student Solutions Manual, Single Variable for Calculus: Early Transcendentals

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The average value of a function z = a − x − y over the region R .

Solution Summary: The author explains how to find the average value of a function z=a-x-y over the region R.

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To find: The average value of a function z=a−x−y over the region R.

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Chapter 13 Solutions