Explanation Let the vector field be F → = x i ^ + 2 y j ^ + z k ^ and S is the triangular region bounded by the intersection of the plane x + 2 y + z = 1 . So, f ( x , y ) = z = − x − 2 y + 1 And, f x = − 1 and f y = − 2 . 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 2 〈 x , 2 y ,

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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 34PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+2yj^+zk^ and S is the triangular region bounded by the intersection of the plane x+2y+z=1 with the positive coordinate planes.

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Chapter 13.5, Problem 33PS

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x i ^ + 2 y j ^ + z k ^ and S is the triangular region bounded by the intersection of the plane x + 2 y + z = 1 with the positive coordinate planes.

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=xi^+2yj^+zk^ and S is the triangular region bounded by the intersection of the plane x+2y+z=1 with the positive coordinate planes.

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Chapter 13 Solutions