Explanation Given information: The figure is given as The domain of the ceiling is defined as 0 ≤ x ≤ π and 0 ≤ y ≤ 1 . Formula used: In an iterated integral where d y precedes d x , integration is performed first with respect to y ∫ c d ( ∫ a b f ( x , y ) d y ) d x where c ≤ x ≤ d and a ≤ y ≤ b Calculation: The ceiling given covers a volume over the domain of 0 ≤ x ≤ π and 0 ≤ y ≤ 1 . The volume of the function can be determined by a double integration. ∫ 0 1 ∫ 0 π z d x d y = ∫ 0 1 ∫ 0 π y 2 sin x d x d y = ∫ 0 1 y 2 ∫ 0 π sin x d x d y = ∫ 0 1 y 2 [ − cos x ] π 0 d y = ∫ 0 1 y 2 ( − cos π − ( − cos 0 ) ) d y = ∫ 0 1

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 16.2, Problem 53E

To determine

The average height of the ceiling of z=y2sinx in the figure given.

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Chapter 16.2, Problem 52E

Chapter 16.2, Problem 54E

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Chapter 16 Solutions

Calculus - Standalone book

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

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The average height of the ceiling of z = y 2 sin x in the figure given.

Solution Summary: The figure is given as the average height of the ceiling of z=y2mathrmsinx in the given figure.

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The average height of the ceiling of z=y2sinx in the figure given.

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Chapter 16 Solutions