Explanation Let the vector field be F → = x 2 i ^ + z k ^ and S is the parametric surface x = sin u cos v , y = sin u sin v , z = cos u for 0 ≤ u ≤ π , 0 ≤ v ≤ 2 π . F ( u , v ) = ( sin u cos v ) 2 i ^ + ( cos u ) k ^ We have, R ( u , v ) = ( sin u cos v ) i ^ + ( sin u sin v ) j ^ + ( cos u ) k ^ R u = ( cos u cos v ) i ^ + ( cos u sin v ) j ^ + ( − sin u ) k ^ And, R v = ( − sin u sin v ) i ^ + ( sin u cos v ) j ^ + ( 0 ) k ^ So, R u × R v = | i j k cos u cos v cos u sin v − sin u − sin u sin v sin u cos v 0 | = ( sin 2 u cos v ) i ^ − ( − sin 2 u sin v ) j ^ + ( 2 sin u cos u cos 2 v ) k ^ So, the value of the integral ∬ S F → ⋅ N → d S is, ∬ S F → ⋅ N → d S = ∬ D F →<

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ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 47PS

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=x2i^+zk^ and S is the parametric surface x=sinucosv , y=sinusinv , z=cosu for 0≤u≤π, 0≤v≤2π .

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Chapter 13.5, Problem 46PS

Chapter 13.5, Problem 48PS

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Chapter 13 Solutions

Calculus

Ch. 13.1 - Prob. 1PS Ch. 13.1 - Prob. 2PS Ch. 13.1 - Prob. 3PS Ch. 13.1 - Prob. 4PS Ch. 13.1 - Prob. 5PS Ch. 13.1 - Prob. 6PS Ch. 13.1 - Prob. 7PS Ch. 13.1 - Prob. 8PS Ch. 13.1 - Prob. 9PS Ch. 13.1 - Prob. 10PS

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The value of the integral ∬ S F → ⋅ N → d S for the vector field F → = x 2 i ^ + z k ^ and S is the parametric surface x = sin u cos v , y = sin u sin v , z = cos u for 0 ≤ u ≤ π , 0 ≤ v ≤ 2 π .

Solution Summary: The author calculates the value of the integral displaystyle

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To calculate: The value of the integral ∬SF→⋅N→dS for the vector field F→=x2i^+zk^ and S is the parametric surface x=sinucosv , y=sinusinv , z=cosu for 0≤u≤π, 0≤v≤2π .

Want to see the full answer?

Chapter 13 Solutions