Explanation Given: The region D shown in the given figure is the union of a type I and type II region. Calculation: Consider if f is continuous and f ( x , y ) ≥ 0 on the region D, the volume of the solid under the surface z = f ( x , y ) ,above the region D is written as follows: V = ∬ D f ( x , y ) d A In this problem, it is required to evaluate a double integral over the union of two regions say D 1 And D 2 .Then the integral can be written as ∬ R ( x y ) d A = ∬ D 1 ( x y ) d A + ∬ D 2 ( x y ) d A Here D 1 = { ( x , y ) : x 2 + y 2 = 1 , y ≥ 0 } and D 2 = { ( x , y ) : − 1 ≤ x < 1 , − x 3 − 1 3 ≤ y ≤ x 2 − 1 } Take y and x respectively as the first variables in the order of integration for integrates over D 1 and D 2 ∬ D 1 ( x y ) d y d x + ∬ D 2

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 12.2, Problem 60PS

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To find: Evaluate ∬Dxy dA

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Chapter 12.2, Problem 59PS

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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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Evaluate ∬ D x y d A

Solution Summary: The author explains how the double integral evaluates to an approximate value of -0.75units squared.

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Chapter 12 Solutions