To derive : The integral theorem: ∫ volume τ ∇ ϕ d τ = ∫ surface enclosing τ ϕ n d τ .

Question

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Answer 1

Question

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

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2. Use Broyden's method to solve the nonlinear system x² + y²+sin(x + y) x+y+cos(xy) = a = = 5 where a is the number formed by the first two digits of your SUID. Try at least two different starting points (or more, if needed to find a solution). Report the outcome of running the method for each starting point. Was the root you found always the same?

The correct answer is Ccould you show me how to do it by finding a0 and and akas well as setting up the piecewise function and integrating

This is an example only. What can be a simialr equation with differnet numbers using logs and can have a mistake in one of the steps and what will be the correct way to solve it. Thanks

Answer 2

Question

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Expert Solution & Answer

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See solution

Answer 4

Question

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Answer 6

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

Answer 7

Chapter 6.11, Problem 17P

a.

To determine

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

Answer 8

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

Answer 9

b.

To determine

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

Answer 10

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

Answer 11

c.

To determine

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

Answer 12

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

Answer 13

d.

To determine

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

Answer 14

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

Answer 15

e.

To determine

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

Answer 16

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

Answer 17

f.

To determine

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

Answer 18

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Answer 19

g.

To determine

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

Answer 20

Expert Solution & Answer

Answer 21

Expert Solution & Answer

Answer 22

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Answer 23

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To derive : The integral theorem: ∫ volume τ ∇ ϕ d τ = ∫ surface enclosing τ ϕ n d τ .

Solution Summary: The Gauss divergence theorem states that if fis a continuously differentiable vector point function and V is the volume bounded by the surface Sthen.

Concept explainers

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

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Chapter 6 Solutions

To derive : The integral theorem: ∫ volume τ ∇ ϕ d τ = ∫ surface enclosing τ ϕ n d τ .

Solution Summary: The Gauss divergence theorem states that if fis a continuously differentiable vector point function and V is the volume bounded by the surface Sthen.

Concept explainers

To derive: The integral theorem: ∫volume τ∇ϕdτ=∫surface enclosing τϕ n dτ .

To derive: The integral theorem: ∫volume τ∇×Vdτ=∫surface enclosing τn× Vdτ

To derive: The integral theorem: ∫curve bounding σϕdr=∫surface σ(n× ∇ϕ)dσ

To derive: The integral theorem: ∮C= curve bounding σdr× V=∫surface σ(n×∇)× Vdσ

To derive: The integral theorem: ∫volume τϕ∇⋅Vdτ=∮surface enclosing τ ϕV⋅ndσ-∫volume τ V⋅∇ϕdτ

To derive: The integral theorem: ∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

To derive: The integral theorem: ∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr

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Chapter 6 Solutions

To derive: The integral theorem:

∫volume τV⋅(∇×U)dτ=∫volume τ U⋅(∇×V)dτ+∮surface enclosing τ (U×V)⋅ndσ

To derive: The integral theorem:

∫surface of σϕ(∇×V)⋅ndσ=∫surface of σ (∇×∇ϕ)⋅ndσ+∮curve bounding σ (ϕV)⋅dr