Explanation Given: The integral: ∫ 0 1 ∫ − x − x 2 x − x 2 ( x 2 + y 2 ) d y d x Formula Used: The following formula of multiple integration is used in order to find the value of the iterated integral: ∬ f ( x ) f ( y ) d x d y = ∫ ( ∫ f ( x ) f ( y ) d x ) d y Calculation: Convert the double integral into polar coordinates by substituting: x = r cos θ y = r sin θ d x d y = r d r d θ The limits of the provided double integral are: 0 ≤ x ≤ 1 − x − x 2 ≤ y ≤ x − x 2 The region bounded by these curves is shown below, Since y = x − x 2 , y = − x − x 2 , x = 0 , x = 1 , y 2 = x − x 2 r 2 sin 2 θ = r cos θ − r 2 cos 2 θ r 2 sin 2 θ + r 2 cos 2 θ = r cos θ r 2 = r cos θ On further simplification:, r 2 − r cos θ = 0 r ( r − cos θ ) = 0 r = 0 , r = cos θ As x = 0 r cos θ = 0 c o s θ × cos θ = 0 cos 2 θ = 0 cos θ = 0 θ = π 2 The corresponding limits in polar coordinates will be: − π 2 ≤ θ ≤ π 2 0 ≤ r ≤ cos θ It should be noted that the limits can also be found from the figure above

WebAssign Printed Access Card for Larson/Edwards' Calculus, Multi-Term

11th Edition

ISBN: 9781337652650

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.3, Problem 20E

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Chapter 14.3, Problem 19E

Chapter 14.3, Problem 21E

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Chapter 14 Solutions

WebAssign Printed Access Card for Larson/Edwards' Calculus, Multi-Term

Ch. 14.1 - CONCEPT CHECK Iterated Integral Explain what is...Ch. 14.1 - Prob. 2E Ch. 14.1 - Question: Evaluate the integral: 0x(2xy)dy Ch. 14.1 - Evaluating an IntegralIn Exercises 310, evaluate...Ch. 14.1 - Evaluating an IntegralIn Exercises 310, evaluate...Ch. 14.1 - Evaluating an IntegralIn Exercises 310, evaluate...Ch. 14.1 - Evaluating an Integral In Exercises 3-10, evaluate...Ch. 14.1 - Evaluating an IntegralIn Exercises 310, evaluate...Ch. 14.1 - Evaluating an Integral In Exercises 3-10, evaluate...Ch. 14.1 - Evaluating an Integral In Exercises 3-10, evaluate...

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What the value of the iterated integral by converting to polar coordinates will be.

Solution Summary: The author explains how to calculate the value of the iterated integral by converting it to polar coordinates.

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Chapter 14 Solutions