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 = ( z , z , x ) z = 9 - x 2 − y 2 a n d x , y , z ≥ 0. Formula used: ∬ s F . d S = ∬ R F ( G ( x , y ) . N ( x , y ) d x d y Calculation: let By parameterizing x = x , y = y , z = 9 − x 2 − y 2 w i t h R = { ( x , y ) | x 2 + y 2 ≤ 9 , x , y ≥ 0. G ( x , y ) = ( x , y , 9 − x 2 − y 2 ) . Now G ( x , y ) = ( x , y , 9 − x 2 − y 2 ) . then T x ( x , y ) = ( 1 , 0 , − 2 x ) T y ( x , y ) = ( 0 , 1 , − 2 y ) Then normal vector is N ( x , y ) = T x ( x , y ) x T y ( x , y ) = | i j k 1 0 − 2 x 0 1 − 2 y | = ( 2 x , 2 y , 1 ) so the normal as N ( x , y ) = ( 2 x , 2 y , 1 ) N o w F ( x , y ) . N ( x , y ) = ( x , y , 9 − x 2 − y 2 ) ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 17.5, Problem 9E

To determine

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

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Chapter 17.5, Problem 8E

Chapter 17.5, Problem 10E

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Chapter 17 Solutions

CALCULUS (CLOTH)

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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The surface integral for given vector function and surface.

Solution Summary: The author explains how to calculate the surface integral for given vector function and surface.

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

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Chapter 17 Solutions