Explanation Formula used: The iterated integral is evaluated as ∬ R f ( x , y ) d A = ∫ c d ( ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ) d y , where f is a continuous function and as well as h 1 and h 2 are continuous defined as h 1 ( y ) ≤ h 2 ( y ) on the interval [ c , d ] and R = { ( x , y ) | h 1 ( y ) ≤ x ≤ h 2 ( y ) , c ≤ y ≤ d } . Calculation: Use online graphing calculator to sketch the region bounded by the lines x = 0 , x = y and y = 4 . From Figure 1, it is observed that the intersection point are ( 2 , 4 ) . The limits are 0 ≤ x ≤ y and 0 ≤ y ≤ 4 . Substitute the values 0 ≤ x ≤ y , 0 ≤ y ≤ 4 and f ( x , y ) = x e − y 2 in the above formula ∬ R f ( x , y ) d A = ∫ c d ( ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ) d y , That is, ∬ R f ( x , y ) d A = ∫ 0 4 ( ∫ 0 y x e − y 2 d x ) d y

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

10th Edition

ISBN: 9781285464640

Author: Soo T. Tan

Publisher: Brooks Cole

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Chapter 8.6, Problem 22E

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To evaluate: The double integral ∬Rf(x,y)dA where f(x,y)=xe−y2 , R is bounded by the lines x=0 , x=y and y=4 .

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Chapter 8.6, Problem 21E

Chapter 8.6, Problem 23E

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Evaluate f(x, y, z) dV for the function f(x, y, z) = xz² and region B = [0, 3] × [5, 7] × [4, 5].

Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

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The double integral ∬ R f ( x , y ) d A where f ( x , y ) = x e − y 2 , R is bounded by the lines x = 0 , x = y and y = 4 .

Solution Summary: The author evaluates the value of the iterated integral, which is 14e16, using an online graphing calculator.

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To evaluate: The double integral ∬Rf(x,y)dA where f(x,y)=xe−y2 , R is bounded by the lines x=0 , x=y and y=4 .

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Chapter 8 Solutions