Explanation Given Info: Given G ( u , v ) = ( u cos v , u sin v , u ) , 0 ≤ u ≤ 1 , 0 ≤ v ≤ 1 . f ( x , y , z ) = z ( x 2 + y 2 ) . Formula Used: Let G ( u , v ) be a parametrization of a surface S with parameter domain D . Assume that G is continuously differentiable, one-to-one, and regular (except possibly at the boundary of D ). Then, ∬ S f ( x , y , z ) d S = ∬ D f ( G ( u , v ) ) ‖ N ( u , v ) ‖ d u d v . Where N ( u , v ) = T u × T v , T u = ∂ G ∂ u , T v = ∂ G ∂ v Calculation: T u = ∂ ( u cos v , u sin v , u ) ∂ u T u = ( cos v , sin v , 1 ) T v = ∂ ( u cos v , u sin v , u ) ∂ v T v = ( − u sin v , u cos v , 0 ) N ( u , v ) = i j k cos v sin v 1 − u sin v u cos v 0 N ( u , v ) = − u cos v i − u sin v j + u k The normal vector has length, ‖ N ( u , v ) ‖ = ( − u cos v ) 2 + ( − u sin v ) 2 + u 2 ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17.4, Problem 13E

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To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.

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Chapter 17.4, Problem 12E

Chapter 17.4, Problem 14E

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Chapter 17 Solutions

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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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∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author explains how to calculate S f (x, y, and z) d S for the given surface and function.

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To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.

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Chapter 17 Solutions

∬ S f ( x , y , z ) d S for the given surface and function.

Solution Summary: The author explains how to calculate S f (x, y, and z) d S for the given surface and function.

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To calculate: ∬Sf(x,y,z)dS for the given surface and function.

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Chapter 17 Solutions

To calculate: $\iint_{S} f (x, y, z) d S$ for the given surface and function.