Explanation Given: ∫ 0 1 ∫ x 1 + x ∫ 0 x y y − 1 z d z d y d x Concept Used: ∫ x n d x = x n + 1 n + 1 + C , where C is an arbitrary constant Calculation: To solve the iterated triple integrals ∫ 0 1 ∫ x 1 + x ∫ 0 x y y − 1 z d z d y d x First, evaluate the inner integral with respect to z treating x and y as constant, as ∫ 0 1 ∫ x 1 + x ∫ 0 x y y − 1 z d z d y d x = ∫ 0 1 ∫ x 1 + x y − 1 [ ∫ 0 x y z d z ] d y d x = ∫ 0 1 ∫ x 1 + x y − 1 [ z 2 2 ] 0 x y d y d x = ∫ 0 1 ∫ x 1 + x y − 1 [ ( x y ) 2 − 0 2 2 ] d y d x = 1 2 ∫ 0 1 ∫ x 1 + x y − 1 ( x y ) 2 d y d x = 1 2 "

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 12.5, Problem 12PS

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To solve the iterated triple integrals

∫01∫ x 1+x ∫ 0 xy y −1 zdzdydx

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Chapter 12.5, Problem 11PS

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Chapter 12 Solutions

Student's Solution and Survival Manual for Calculus

Ch. 12.1 - Prob. 1PS Ch. 12.1 - Prob. 2PS Ch. 12.1 - Prob. 3PS Ch. 12.1 - Prob. 4PS Ch. 12.1 - Prob. 5PS Ch. 12.1 - Prob. 6PS Ch. 12.1 - Prob. 7PS Ch. 12.1 - Prob. 8PS Ch. 12.1 - Prob. 9PS Ch. 12.1 - Prob. 10PS

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To solve the iterated triple integrals ∫ 0 1 ∫ x 1 + x ∫ 0 x y y − 1 z d z d y d x

Solution Summary: The author evaluates the inner integral with respect to z treating x and y as constants.

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To solve the iterated triple integrals ∫01∫ x 1+x ∫ 0 xy y −1 zdzdydx

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Chapter 12 Solutions

To solve the iterated triple integrals

∫01∫ x 1+x ∫ 0 xy y −1 zdzdydx