Explanation Given: f ( x , y ) = x y x 2 + y 2 + 1 ; R : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 Calculation: Consider a line f ( x , y ) f ( x , y ) = x y x 2 + y 2 + 1 And rectangle R such that R : 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 Find the volume of solid bounded below by rectangle R in the x y plane and above by the graph of z = f ( x , y ) So volume will be evaluated by finding the following integral ∫ ∫ R f ( x , y ) d A This is same as; ∫ ∫ R f ( x , y ) d A = ∫ 0 1 ∫ 0 1 ( x y x 2 + y 2 + 1 ) d y d x So evaluate ∫ 0 1 ∫ 0 1 ( x y x 2 + y 2 + 1 ) d y d x First keeping x as constant integrate with respect to y ∫ ∫ R f ( x , y ) d A = ∫ 0 1 ∫ 0 1 ( x y x 2 + y 2 + 1 ) d y d x = ∫ 0 1 ( ∫ 0 1 x y x 2 + y 2 + 1 d y ) d x Since ∫ c b ∫ c d f ( x , y ) d y d x = ∫ c b ( ∫ c d f ( x , y ) d y d x ) Further solving Let t = y 2 , differentiate t with respect to y d t d y = 2 y 2 y d y = d t y d y = d t 2 And when y = 0 then t = 0 and when y = 1 then t = 1 So the integral ∫ ∫ R f ( x , y ) d A = ∫ 0 1 ∫ 0 1 ( x y x 2 + y 2 + 1 ) d y d x becomes ∫ ∫ R f ( x , y ) d A = 1 2 ∫ 0 1 ( ∫ 0 1 x x 2 + y 2 + 1 d t d x ) = 1 2 ∫ 0 1 ( x x

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

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

Publisher: Kendall Hunt Publishing

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Chapter 12.1, Problem 37PS

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To find: the volume of the solid bounded.

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Chapter 12.1, Problem 36PS

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Chapter 12 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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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the volume of the solid bounded.

Solution Summary: The author explains how to find the volume of the solid bounded below by rectangle R in the xy plane and above by the graph.

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