Explanation Let the vector field be F → = x 2 i ^ + y 2 j ^ + z 2 k ^ and S is the portion of the plane z = y + 1 that lies inside the cylinder x 2 + y 2 = 1 . So, f ( x , y ) = z = y + 1 And, f x = 0 and f y = 1 . So, the value of the integral ∬ S F → ⋅ N → d S is, ∬ S F → ⋅ N → d S = ∬ x 2 + y 2 = 1 F → ( x , y , f ( x , y ) ) ⋅ 〈 − f x , − f y , 1 〉 d A = ∫ x = − 1 1 ∫ y = − 1 1 〈 x 2 , <

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 37PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=x2i^+y2j^+z2k^ and S is the portion of the plane z=y+1 that lies inside the cylinder x2+y2=1 .

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Chapter 13.5, Problem 36PS

Chapter 13.5, Problem 38PS

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Chapter 13 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x 2 i ^ + y 2 j ^ + z 2 k ^ and S is the portion of the plane z = y + 1 that lies inside the cylinder x 2 + y 2 = 1 .

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=x2i^+y2j^+z2k^ and S is the portion of the plane z=y+1 that lies inside the cylinder x2+y2=1 .

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Chapter 13 Solutions