Explanation Given: Given integral is ∬ R ( 4 − x 2 − y 2 ) d A where R is the region over the rectangle: 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 Concept used: When a double integral of constant limits is given, the integral can be evaluated by taking any order of integration . Calculation: Limits of x : 0 to 1 Limits of y : 0 to 1 ⇒ ∫ 0 1 ∫ 0 1 ( 4 − x 2 − y 2 ) d y d x ⇒ ∫ 0 1 ( ∫ 0 1 ( 4 − x 2 − y 2 ) d y ) d x ⇒ ∫ 0 1 ( [ 4 y − x 2 y − y 3 3 ] 0 1 ) d x ⇒ ∫ 0 1 ( [ 4 ( 1 ) − x 2 ( 1 ) − ( 1 ) 3 3 ] −

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Question

Chapter 12.1, Problem 51PS

To determine

To evaluate: ∬R(4−x2−y2) dA

Where R is the region over the rectangle:

0≤x≤1,0≤y≤1

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Chapter 12.1, Problem 50PS

Chapter 12.1, Problem 52PS

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Chapter 12 Solutions

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

Knowledge Booster

∬ R ( 4 − x 2 − y 2 ) d A Where R is the region over the rectangle: 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1

Solution Summary: The author explains how a double integral of constant limits can be evaluated by taking any order of integration.

To evaluate: ∬R(4−x2−y2) dA

Where R is the region over the rectangle:

0≤x≤1,0≤y≤1

Want to see the full answer?

Chapter 12 Solutions

∬ R ( 4 − x 2 − y 2 ) d A Where R is the region over the rectangle: 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1

Solution Summary: The author explains how a double integral of constant limits can be evaluated by taking any order of integration.

To evaluate: ∬R(4−x2−y2) dA Where R is the region over the rectangle: 0≤x≤1,0≤y≤1

Want to see the full answer?

Chapter 12 Solutions

To evaluate: ∬R(4−x2−y2) dA

Where R is the region over the rectangle:

0≤x≤1,0≤y≤1