To determine Find the surface integral of the field F ( x , y , z ) = z k over the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin. Explanation Given information: The field F ( x , y , z ) = z k over the portion of the sphere x 2 + y 2 + z 2 = a 2 in the first octant in the direction away from the origin. Definition: If S is part of a level surface g ( x , y , z ) = c , then n may be taken to be one of the two field n = ± ∇ g | ∇ g | , depending on which one gives the preferred direction. The corresponding flux is ∬ S F ⋅ n d σ = ∬ R ( F ⋅ ± ∇ g | ∇ g | ) | ∇ g | | ∇ g ⋅ p | d A = ∬ R F ⋅ ± ∇ g | ∇ g ⋅ p | d A Calculation: Write the equation. g : x 2 + y 2 + z 2 = a 2 (1) Differentiate Equation (1). ∇ g = 2 x i + 2 y j + 2 z k Find determinant of | ∇ g | . | ∇ g | = ( 2 x ) 2 + ( 2 y ) 2 + ( 2 z ) 2 = 4 x 2 + 4 y 2 + 4 z 2 = 2 x 2 + y 2 + z 2 = 2 a Find n = ∇ g | ∇ g | . n = ∇ g | ∇ g | = 2 x i + 2 y j + 2 z k 2 a = x i + y j + z k a To get a unit vector normal to the plane of R , take p = k . Find | ∇ g ⋅ k | . | ∇ g ⋅ k | = | ( 2 x i + 2 y j + 2 z k ) ⋅ ( k ) | = | 2 z | = ( 2 z ) 2 = 2 z Find d σ = | ∇ g | | ∇ g ⋅ k | d A

14th Edition

ISBN: 9780134438986

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

Publisher: PEARSON

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Chapter 16.6, Problem 31E

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Find the surface integral of the field F(x,y,z)=zk over the portion of the sphere x2+y2+z2=a2 in the first octant in the direction away from the origin.

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Chapter 16.6, Problem 30E

Chapter 16.6, Problem 32E

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Thomas' Calculus (14th Edition)

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