Explanation Given: F ( x , y , z ) = 〈 0 , 0 , x z e x y 〉 , z = x y , z ≤ 0 , 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 with upward-pointing normal. Formula used: ∬ S F . d S = ∬ S F . N d S Calculation: Parametrization of given surface is G ( u , v ) = ( u , v , u v ) , 0 ≤ u ≤ 1 , 0 ≤ v ≤ 1 G u ( u , v ) = ( 1 , 0 , v ) , G v ( u , v ) = ( 0 , 1 , u ) Thus, N= G u × G v = | i j k 1 0 v 0 1 u | = i ( − v ) + j ( − u ) + k ( 1 ) = 〈 − v , − u , 1 〉 Hence, the normal vector N = 〈 − v , − u , 1 〉 Therefore, ∬ S F . d S = ∫ 0 1 ∫ 0 1 〈 0 , 0 , u 2 v e u v 〉 · 〈 − v , − u , 1 〉 d u d v = ∫ 0 1 ∫ 0 1 u 2

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17, Problem 56CRE

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The value of the integral $\iint_{S} F . d S$ .

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Chapter 17, Problem 55CRE

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Chapter 17 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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The value of the integral ∬ S F . d S .

Solution Summary: The author explains the formula used to determine the value of the integral S F. d S.

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The value of the integral $\iint_{S} F . d S$ .

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Chapter 17 Solutions

The value of the integral ∬ S F . d S .

Solution Summary: The author explains the formula used to determine the value of the integral S F. d S.

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The value of the integral ∬SF.dS.

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Chapter 17 Solutions

The value of the integral $\iint_{S} F . d S$ .