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 bellow. F = ( 0 , 3 , x ) a n d x 2 + y 2 + z 2 = 1 a n d x , y , z ≥ 0. Formula Used: ∬ s F . d S = ∬ R F ( G ( r , s ) . N ( r , s ) d r d s Calculation: let By parameterizing x = r cos t , y = r sin t , z = 1 − r 2 w i t h 0 ≤ r ≤ 3 , 0 ≤ t ≤ π 2 . G ( r , t ) = ( r cos t , r sin t , 1 − r 2 ) w i t h ; 0 ≤ r ≤ 3 , 0 ≤ t ≤ π 2 . Now G ( r , t ) = ( r cos t , r sin t , 1 − r 2 ) then T r ( r , t ) = ( cos t , sin t , − r 1 − r 2 ) T t ( r , t ) = ( − r sin t , r cos t , 0 ) Then normal vector is N ( r , t ) = T r ( r , t ) x T t ( r , t ) = | i j k cos t sin t − r 1 − r 2 − r sin t r cos t 0 | = i ( r 2 cost 1 − r 2 ) − j ( − r 2 s i n t 1 ...

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.5, Problem 7E

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

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Chapter 17.5, Problem 6E

Chapter 17.5, Problem 8E

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Chapter 17 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 17.1 - Prob. 1PQ Ch. 17.1 - Prob. 2PQ Ch. 17.1 - Prob. 3PQ Ch. 17.1 - Prob. 4PQ Ch. 17.1 - Prob. 1E Ch. 17.1 - Prob. 2E Ch. 17.1 - Prob. 3E Ch. 17.1 - Prob. 4E Ch. 17.1 - Prob. 5E Ch. 17.1 - Prob. 6E

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Thesurface integral for given vector function and surface.

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Chapter 17 Solutions