Explanation Given: Velocityfield is v = 〈 2 x , y , x y 〉 m / s across the part of the cylinder x 2 4 + y 2 = 1 , x ≥ 0 , y ≥ 0 , 0 ≤ z ≤ 4 Formula used: ∬ S F . d S = ∬ S F . N d S Calculation: Parametrization of given cylinder is S ( θ , z ) = ( 2 cos θ , sin θ , z ) , 0 ≤ θ ≤ π 2 , 0 ≤ z ≤ 4 S θ ( θ , z ) = ( − 2 sin θ , cos θ , 0 ) S z ( θ , z ) = ( 0 , 0 , 1 ) Thus, N= S θ × S z = | i j k − 2 sin θ cos θ 0 0 0 1 | = i ( cos θ ) + j ( 2 sin θ ) + k ( 0 ) = 〈 cos θ , 2 sin θ , 0 〉 Hence, the normal vector N = 〈 cos θ , 2 sin θ , 0 〉 Therefore, Flow rate of fluid = ∬ S F ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 17, Problem 60CRE

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The flow rate of a fluid.

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Chapter 17, Problem 59CRE

Chapter 17, Problem 61CRE

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Chapter 17 Solutions

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The flow rate of a fluid.

Solution Summary: The author calculates the flow rate of a fluid using the following formula: d S = S F.

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Chapter 17 Solutions