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 = 4 − y 2 , y = 0 . From Figure 1, it is observed that the intersection point are ( 0 , 2 ) and the shaded region is the required region. Therefore, the limit of integration will be 0 ≤ x ≤ 4 − y 2 and 0 ≤ y ≤ 2 . Substitute the values 0 ≤ x ≤ 4 − y 2 , 0 ≤ y ≤ 2 and f ( x , y ) = y 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 , ∬ R f ( x , y ) d A = ∫ 0 2 ( ∫ 0 4 − y 2 y d x ) d y The first integral

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 8.7, Problem 17E

To determine

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

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Chapter 8.7, Problem 16E

Chapter 8.7, Problem 18E

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Chapter 8 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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

Solution Summary: The author evaluates the iterated integral by using an online graphing calculator. The intersection point is (0,2) and the shaded region is the required region.

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

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Chapter 8 Solutions