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 . Δ is Laplace operator. 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 directional derivative is defined as: D n ϕ = 〈 ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z 〉 ⋅ n Let us consider F ⋅ n = D n ϕ = 〈 ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z 〉 ⋅ n ... To determine (b) To prove: If ϕ is a harmonic function then: ∬ S D n ϕ d S = 0

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ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 18.3, Problem 39E

To determine

(a)

To prove:

$\iint_{S} D_{n} ϕ d S = ∭_{W} △ ϕ d V$

To determine

(b)

To prove:

If $ϕ$ is a harmonic function then: $\iint_{S} D_{n} ϕ d S = 0$

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Chapter 18.3, Problem 38E

Chapter 18.3, Problem 40E

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Chapter 18 Solutions

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(a) To prove: ∬ S D n ϕ d S = ∭ W △ ϕ d V

Solution Summary: The author proves that directional derivative is defined as: D n = outward unit normal vector and Laplace operator.

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(a)

To prove:

$\iint_{S} D_{n} ϕ d S = ∭_{W} △ ϕ d V$

(b)

To prove:

If $ϕ$ is a harmonic function then: $\iint_{S} D_{n} ϕ d S = 0$

Want to see the full answer?

Chapter 18 Solutions

(a) To prove: ∬ S D n ϕ d S = ∭ W △ ϕ d V

Solution Summary: The author proves that directional derivative is defined as: D n = outward unit normal vector and Laplace operator.

Concept explainers

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(a) To prove: ∬SDnϕdS=∭W△ϕdV

(b) To prove: If ϕ is a harmonic function then: ∬SDnϕdS=0

Want to see the full answer?

Chapter 18 Solutions

(a)

To prove:

$\iint_{S} D_{n} ϕ d S = ∭_{W} △ ϕ d V$

(b)

To prove:

If $ϕ$ is a harmonic function then: $\iint_{S} D_{n} ϕ d S = 0$