Explanation Let the vector field be F → = x i ^ + 2 y j ^ − 3 z k ^ and S is that part of the plane 15 x − 12 y + 3 z = 6 . So, f ( x , y ) = z = − 5 x + 4 y + 2 And, f x = − 5 and f y = 4 . 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 ∫ y = 0 1 〈 x , 2 y , 15 x − 12 y − 6 〉 ⋅ 〈 − ( − 5 ) , −

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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 33PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+2yj^−3zk^ and S is that part of the plane 15x−12y+3z=6 that lies above the unit square 0≤x≤1, 0≤y≤1 .

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Chapter 13.5, Problem 32PS

Chapter 13.5, Problem 34PS

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Chapter 13 Solutions

Calculus

Ch. 13.1 - Prob. 1PS Ch. 13.1 - Prob. 2PS Ch. 13.1 - Prob. 3PS Ch. 13.1 - Prob. 4PS Ch. 13.1 - Prob. 5PS Ch. 13.1 - Prob. 6PS Ch. 13.1 - Prob. 7PS Ch. 13.1 - Prob. 8PS Ch. 13.1 - Prob. 9PS Ch. 13.1 - Prob. 10PS

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x i ^ + 2 y j ^ − 3 z k ^ and S is that part of the plane 15 x − 12 y + 3 z = 6 that lies above the unit square 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 .

Solution Summary: The author calculates the value of the integral displaystyle

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+2yj^−3zk^ and S is that part of the plane 15x−12y+3z=6 that lies above the unit square 0≤x≤1, 0≤y≤1 .

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Chapter 13 Solutions