Explanation Let the vector field be F → = x i ^ + y j ^ + 2 z k ^ and S is the surface of the cube bounded by the planes x = 0 , x = 1 , y = 0 , y = 1 , z = 0 and z = 1 . Here, S is the surface of the cube bounded by the planes x = 0 , x = 1 , y = 0 , y = 1 , z = 0 and z = 1 , that is, x + y + z = 1 . So, f ( x , y ) = z = − x − y + 1 And, f x = − 1 and f y = − 1 . 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 = 0 1

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 36PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+yj^+2zk^ and S is the surface of the cube bounded by the planes x=0,x=1,y=0,y=1,z=0 and z=1 .

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Chapter 13.5, Problem 35PS

Chapter 13.5, Problem 37PS

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Chapter 13 Solutions

Student's Solution and Survival Manual for Calculus

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x i ^ + y j ^ + 2 z k ^ and S is the surface of the cube bounded by the planes x = 0 , x = 1 , y = 0 , y = 1 , z = 0 and z = 1 .

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+yj^+2zk^ and S is the surface of the cube bounded by the planes x=0,x=1,y=0,y=1,z=0 and z=1 .

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Chapter 13 Solutions