Explanation Given: The vector field F ( x , y , z ) = 〈 x y z + x y , 1 2 y 2 ( 1 − z ) + e x , e x 2 + y 2 〉 . The surface S is the boundary the cylinder x 2 + y 2 = 16 ; z = 0 , z = y − 4 Formula used: Divergence Theorem, ∬ S F ⋅ d S = ∭ E d i v F d V Calculation: Let us first find the divergence of the given vector field. ⇒ F ( x , y , z ) = 〈 x y z + x y , 1 2 y 2 ( 1 − z ) + e x , e x 2 + y 2 〉 ⇒ d i v F = 〈 ∂ ∂ x , ∂ ∂ y , ∂ ∂ z 〉 ⋅ 〈 x y z + x y , 1 2 y 2 ( 1 − z ) + e x , e x 2 + y 2 〉 ⇒ d i v F = ∂ ∂ x ( x y z + x y ) + ∂ ∂ y ( 1 2 y 2 ( 1 − z ) + e x ) + ∂ ∂ z ( e x 2 + y 2 ) ⇒ d i v F = y z + y + y ( 1 − z ) + 0 = y z + y + y − y z = 2 y Using divergence theorem, we can convert the surface integral to volume integral as shown below: ⇒ ∬ S F ⋅ d S = ∭ E d i v F d V ⇒ ∬ S F ⋅ d S = ∭ E ( 2 y ) d V For the given cylinder x 2 + y 2

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18, Problem 25CRE

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The surface integral $\iint_{S} F \cdot d S$ using Divergence Theorem.

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Chapter 18, Problem 24CRE

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The surface integral ∬ S F ⋅ d S using Divergence Theorem.

Solution Summary: The author evaluates the surface integral using the divergence theorem.

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The surface integral $\iint_{S} F \cdot d S$ using Divergence Theorem.

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Chapter 18 Solutions

The surface integral ∬ S F ⋅ d S using Divergence Theorem.

Solution Summary: The author evaluates the surface integral using the divergence theorem.

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To evaluate: The surface integral ∬SF⋅dS using Divergence Theorem.

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Chapter 18 Solutions

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The surface integral $\iint_{S} F \cdot d S$ using Divergence Theorem.