Explanation Let the vector field be F → = y i ^ + z 2 j ^ − 2 z k ^ and S is the part of the surface cut from the cylinder z = 3 − x 2 by y = 0 , y = 1 , and z = 0 . So, f ( x , y ) = z = 3 − x 2 And, f x = − 2 x and f y = 0 . So, the value of the integral ∬ S F → ⋅ N → d S is, ∬ S F → ⋅ N → d S = ∬ D F → ( x , y , f ( x , y ) ) ⋅ 〈 − f x , − f y , 1 〉 d A = ∫ x = − 3 3 ∫ y = 0 1 〈 y , ( 3 − x 2 ) 2 , 2 x 2 − 6 〉 ⋅ 〈 − ( − 2 x ) , − ( 0 )

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 39PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=yi^+z2j^−2zk^ and S is the part of the surface cut from the cylinder z=3−x2 by y=0,y=1 , and z=0 .

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Chapter 13.5, Problem 38PS

Chapter 13.5, Problem 40PS

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = y i ^ + z 2 j ^ − 2 z k ^ and S is the part of the surface cut from the cylinder z = 3 − x 2 by y = 0 , y = 1 , and z = 0 .

Solution Summary: The author calculates the value of the integral displaystyle 'underset' Siint's stackrel to a and S is the surface cut from the cylinder

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=yi^+z2j^−2zk^ and S is the part of the surface cut from the cylinder z=3−x2 by y=0,y=1 , and z=0 .

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Chapter 13 Solutions