Explanation Given information: The surface, S : z = 4 − x , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2 Consider the integral as, ∬ S ( x 2 + y 2 ) d S Let, z = g ( x , y ) Then, g ( x , y ) = 4 − x Now, ∂ g ∂ x = − 1 and ∂ g ∂ y = 0 Then, d S = 1 + ( ∂ g ∂ x ) 2 + ( ∂ g ∂ y ) 2 d A = 1 + ( − 1 ) 2 + ( 0 ) 2 d A = 2 d A Now, the given integral becomes, ∬ S ( x 2 + y 2 ) d S = 2 ∫ x = 0 2 ∫ y = 0 2 ( x 2 + y 2 ) d A = 2 ∫ x = 0 2 ∫ <

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ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 8PS

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To calculate: The value of the integral ∬S( x 2+ y 2)dS where, S:z=4−x,0≤x≤2,0≤y≤2

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Chapter 13.5, Problem 7PS

Chapter 13.5, Problem 9PS

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The value of the integral ∬ S ( x 2 + y 2 ) d S where, S : z = 4 − x , 0 ≤ x ≤ 2 , 0 ≤ y ≤ 2

Solution Summary: The author explains the value of the integral, which is displaystyleundersetSiint.

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To calculate: The value of the integral ∬S( x 2+ y 2)dS where, S:z=4−x,0≤x≤2,0≤y≤2

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Chapter 13 Solutions