Explanation Given: S is the boundary of surface of a region W in R 3 , and D n ϕ is the directional derivative of ϕ , where n is outward unit normal vector . ϕ is harmonic. Formula used: Divergence Theorem: ∬ S F ⋅ n d S = ∭ W D i v ( F ) d V Δ φ = ∂ 2 φ ∂ x 2 + ∂ 2 φ ∂ y 2 + ∂ 2 φ ∂ z 2 Proof: We know that, ⇒ ∇ ϕ = 〈 ∂ ∂ x , ∂ ∂ y , ∂ ∂ z 〉 ϕ ⇒ ∇ ϕ = 〈 ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z 〉 Let us multiply it with ϕ and then take divergence of that product. ⇒ ϕ ∇ ϕ = ϕ 〈 ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z 〉 ⇒ ϕ ∇ ϕ = 〈 ϕ ∂ ϕ ∂ x , ϕ ∂ ϕ ∂ y , ϕ ∂ ϕ ∂ z 〉 Let us now find its divergence: ⇒ d i v ( ϕ ∇ ϕ ) = 〈 ∂ ∂ x , ∂ ∂ y , ∂ ∂ z 〉 ⋅ 〈 ϕ ∂ ϕ ∂ x

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 18.3, Problem 40E

To determine

To prove:

$d i v (ϕ \nabla ϕ) = {‖ \nabla ϕ ‖}^{2}$

And then conclude that,

$\iint_{S} ϕ D_{n} ϕ d S = ∭_{W} {‖ \nabla ϕ ‖}^{2} d V$

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Chapter 18.3, Problem 39E

Chapter 18.3, Problem 41E

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Chapter 18 Solutions

CALCULUS (CLOTH)

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To prove: d i v ( ϕ ∇ ϕ ) = ‖ ∇ ϕ ‖ 2 And then conclude that, ∬ S ϕ D n ϕ d S = ∭ W ‖ ∇ ϕ ‖ 2 d V

Solution Summary: The author proves that d i v is a boundary of surface in R 3 and D n is the directional derivative of.

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To prove:

$d i v (ϕ \nabla ϕ) = {‖ \nabla ϕ ‖}^{2}$

And then conclude that,

$\iint_{S} ϕ D_{n} ϕ d S = ∭_{W} {‖ \nabla ϕ ‖}^{2} d V$

Want to see the full answer?

Chapter 18 Solutions

To prove: d i v ( ϕ ∇ ϕ ) = ‖ ∇ ϕ ‖ 2 And then conclude that, ∬ S ϕ D n ϕ d S = ∭ W ‖ ∇ ϕ ‖ 2 d V

Solution Summary: The author proves that d i v is a boundary of surface in R 3 and D n is the directional derivative of.

Videos

To prove: div(ϕ∇ϕ)=‖∇ϕ‖2 And then conclude that, ∬SϕDnϕdS=∭W‖∇ϕ‖2dV

Want to see the full answer?

Chapter 18 Solutions

To prove:

$d i v (ϕ \nabla ϕ) = {‖ \nabla ϕ ‖}^{2}$

And then conclude that,

$\iint_{S} ϕ D_{n} ϕ d S = ∭_{W} {‖ \nabla ϕ ‖}^{2} d V$