To determine Calculate the flux for the field F = x y i − z k through the portion of the cone z = x 2 + y 2 in the direction away from the z -axis . Explanation Given information: The field F = x y i − z k through the portion of the cone z = x 2 + y 2 , 0 ≤ z ≤ 1 in the direction away from the z -axis . Definition: Let F be a vector field in three-dimensional space with continuous components defined over a smooth surface S having a chosen field of normal unit vectors n orienting S . Then the surface integral of F over S ∬ S F ⋅ n d σ = ∬ R F ⋅ r u × r v | r u × r v | | r u × r v | d u d v = ∬ R F ⋅ ( r u × r v ) d u d v . This integral is also called the flux of the vector field F across S . Calculation: Write the equation. z = x 2 + y 2 (1) F = x y i − z k (2) The parametrization of the Equation (1). r ( r , θ ) = ( r cos θ ) i + ( r sin θ ) j + r k , 0 ≤ r ≤ 1 , 0 ≤ θ ≤ 2 π (3) Differentiate Equation (3) with respect to r . r r ( r , z ) = ( ( 1 ) cos θ ) i + ( ( 1 ) sin θ ) j + ( 1 ) k = ( cos θ ) i + ( sin θ ) j + k Differentiate Equation (3) with respect to θ . r θ ( r , θ ) = ( r ( − sin θ ) ) i + ( r cos θ ) j + ( 0 ) k = ( − r sin θ ) i + ( r cos θ ) j Find r r × r θ . r r × r θ = | i j k − r sin θ r cos θ 0 cos θ sin θ 1 | = [ ( r cos θ ) − ( 0 ) ] i − [ ( − r sin θ ) − ( 0 ) ] j + [ ( − r sin 2 θ ) − ( r cos 2 θ ) ] k = ( r cos θ ) i + ( r sin θ ) j − r ( sin 2 θ + cos 2 θ ) k = ( r cos θ ) i + ( r sin θ ) j − r k Find F ⋅ n d σ

14th Edition

ISBN: 9780134438986

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 16.6, Problem 25E

To determine

Calculate the flux for the field F=xyi−zk through the portion of the cone z=x2+y2 in the direction away from the z-axis.

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Chapter 16.6, Problem 24E

Chapter 16.6, Problem 26E

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Thomas' Calculus (14th Edition)

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