Explanation For funding thesurface integral across the given surface we have to find the surface integral by parameterizing and then evaluate it. Given: The force function and the points are given below. F = ( y , z , 0 ) G ( u , v ) = ( u 3 − v , u + v , v 2 ) ; 0 ≤ u ≤ 2 , 0 ≤ v ≤ 3. F ( u , v ) = ( u + v , v 2 , 0 ) Formula Used: ∬ s F . d S = ∬ R F ( G ( u , v ) . N ( u , v ) d u d v Calculation: Then normal vectors for the surface is G ( u , v ) = ( u 3 − v , u + v , v 2 ) ; T u ( u , v ) = ( 3 u 2 , 1 , 0 ) T v ( u , v ) = ( − 1 , 1 , 2 v ) N ( u , v ) = T u ( u , v ) x T u ( u , v ) = | i j k 3 u 2 1 0 − 1 1 2 v | = ( 2 v , − 6 u 2 v , 3 u 2 + 1 ) Since the normal vector downward so N ( u , v ) = ( − 2 v , 6 u 2 v , − 3 u 2 − 1 ) F ( u , v ) ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.5, Problem 17E

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To calculate: The surface integral for given vector function and surface.

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Chapter 17.5, Problem 16E

Chapter 17.5, Problem 18E

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Chapter 17 Solutions

CALCULUS (CLOTH)

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The surface integral for given vector function and surface.

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Chapter 17 Solutions