Explanation The surface S is defined by R → ( u , v ) = u i ^ − u 2 j ^ + v k ^ , 0 ≤ u ≤ 2 , 0 ≤ v ≤ 1 . Here, x = u , y = − u 2 and z = v . So, f ( R ) = f ( u , − u 2 , v ) = u 2 − u 2 − v = − v Let R → ( u , v ) = u i ^ − u 2 j ^ + v k ^ . R u = 〈 1 , − 2 u , 0 〉 and R v = 〈 0 , 0 , 1 〉 The cross product of R u = 〈 1 , − 2 u , 0 〉 and R v = 〈 0 , 0 , 1 〉 is, R u × R v = | i j k 1 − 2 u 0 0 0 1 | = ( − 2 u ) i ^ − ( 1 ) j ^ + ( 0 ) k ^ So, the magnitude of R u × R v is, ‖ R u × R v ‖ = 4 u 2 + 1 + 0 = 4 u 2 + 1 Hence, the value of the integral ∬ S ( x 2 + y − z )

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

Publisher: Kendall Hunt Publishing

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Chapter 13.5, Problem 45PS

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To calculate: The value of the integral ∬S( x 2+y−z)dS , where S is the surface defined by R→(u,v)=ui^−u2j^+vk^ , 0≤u≤2, 0≤v≤1 .

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Chapter 13.5, Problem 44PS

Chapter 13.5, Problem 46PS

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Chapter 13 Solutions

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The value of the integral ∬ S ( x 2 + y − z ) d S , where S is the surface defined by R → ( u , v ) = u i ^ − u 2 j ^ + v k ^ , 0 ≤ u ≤ 2 , 0 ≤ v ≤ 1 .

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To calculate: The value of the integral ∬S( x 2+y−z)dS , where S is the surface defined by R→(u,v)=ui^−u2j^+vk^ , 0≤u≤2, 0≤v≤1 .

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Chapter 13 Solutions