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9. Let F be any vector field of the form F(x,y, z) f (x) ...

Resolved Question:

Please help quick.


9. Let F be any vector field of the form F(x,y, z) = f (x) i+g(y) j+h(z)k
and let G be any vector field of the form G(x,y, z) = f (y, z) i+g(x, z) j+h(x,y)k.
Indicate whether the following statements are true or false by placing ”T” or ”F” to
the left of the statement.

1. G is incompressible
2. F is incompressible
3. F is irrotational
4. G is irrotational

10.
A vector field gives a geographical description of the flow of money in a society. In the neighborhood of a political convention,
the divergence of this vector field is:
• A. negative
• B. positive
• C. zero

11. Evaluate Double_Integral_S sqrt(1+x^2+y^2) dS where S is the helicoid:
r(u,v) = ucos(v)i+usin(v)j+vk, with 0 <= u <= 1,0 <= v <= 3pi


12. Let S be the part of the plane 3x+3y+z = 4 which
lies in the first octant, oriented upward. Find the flux of the
vector field F = 2i+3j+3k across the surface S.

13. Suppose F is a radial force field, S1 is a sphere of
radius 9 centered at the origin, and the flux integral
Double_Integral_S1 F·dS= 2.
Let S2 be a sphere of radius 36 centered at the origin, and consider
the flux integral
Double_integral_S2 F·dS.

(A) If the magnitude of F is inversely proportional to the
square of the distance from the origin,what is the value of
Double_Integral_S2 F·dS?

(B) If the magnitude of F is inversely proportional to the cube
of the distance from the origin, what is the value of
Double_integral_S2 F·dS?



Please I need the answers as quick as possible
Submitted: 6 years ago.
Category: Homework
Expert:  MrBill31 replied 6 years ago.

Will do this piece meal too so you get answers quicker

9. Let F be any vector field of the form F(x,y, z) = f (x) i+g(y) j+h(z)k
and let G be any vector field of the form G(x,y, z) = f (y, z) i+g(x, z) j+h(x,y)k.
Indicate whether the following statements are true or false by placing "T" or "F" to
the left of the statement.

1. G is incompressible
2. F is incompressible
3. F is irrotational
4. G is irrotational

So if you calculate:

F has a divergence potentially

divF = df/dx + dg/fy + dh/dz

But curlF = 0

(all cross derivatives df/dy df/dz, dg/dx etc = 0

G has zero divergence

divG = df/dx + dg/fy + dh/dz = 0 + 0 + 0 = 0

but it may have a curl

df/dy may be non-zero

df/dz, dg/dx etc may all be non-zero

incompressible = div = 0

irrotational = curl = 0

so

1. G is incompressible

True
2. F is incompressible

False
3. F is irrotational

True
4. G is irrotational

False

10.
A vector field gives a geographical description of the flow of money in a society. In the neighborhood of a political convention,
the divergence of this vector field is:

  • A. negative

  • B. positive
  • C. zero

Divergence is the net flow of money into and out of an area

Unless the politicians have a printing press printing money in the convention, the total supply of money is constant and the net flow has to be zero

answer is C

btw why the dollar has plummeted >50% in value in the last 2 years is because US government DOES have a printing press!

(gutless way to get out of paying its obligations, print more money!)

MrBill31, Engineer
Category: Homework
Satisfied Customers: 85
Experience: BS General Physics and Chemistry MS Physics specialty biophysics
MrBill31 and 9 other Homework Specialists are ready to help you
Customer: replied 6 years ago.
10- C is incorrect
Expert:  MrBill31 replied 6 years ago.

Well then it would be negative (A)

If you look at an instant in time (rather than the total flow over all time) than money flows INTO the convention (divergence would be negative)

Of course at some point the politicians have to leave with their money (in their pockets or whatever) which is why the net flow (averaged over time) must be zero.

Customer: replied 6 years ago.
Correct thank you.
Expert:  MrBill31 replied 6 years ago.

I'm sorry this is the best I can do for now

I can return to this problem at 7 pm

will try to get another expert to help you

Customer: replied 6 years ago.
ok thank you i will be waiting.

The home work has been extended to 10pm tonight.

Thank you.
Expert:  MrBill31 replied 6 years ago.

I'm back,

working...

 

Customer: replied 6 years ago.
Ok i am waiting.

Thank you.
Expert:  MrBill31 replied 6 years ago.

11 confuses me since it is not a closed surface

will work on the rest and then return to it

 

12. Let S be the part of the plane 3x+3y+z = 4 which
lies in the first octant, oriented upward. Find the flux of the
vector field F = 2i+3j+3k across the surface S.

So we have a plane with a unit normal

(1/sqrt(32+32+12)(3i + 3j + k) = (1/sqrt(19))(3i + 3j + k)

it is bounded by

3x+3y+z = 4

x>=0

y>=0

z>=0

which means it is a triangle bounded by the points

(4/3,0,0)

(0,4/3,0)

(0,0,4)

What is the flux per unit area?

FA = n*F = (1/sqrt(19))(3i + 3j + k)*(2i+3j+3k )

= (1/sqrt(19))(3*2+3*3+1*3) = 18/sqrt(19)

what is the area of the triangle?

The base of the triangle is given as the distance between the points

(4/3,0,0) and (0,4/3,0)

b = sqrt(2)*(4/3) = 1.8856181

The mid point on the base is:

(2/3,2/3,0) (average of the points on the base (4/3,0,0) and (0,4/3,0))

the height of the triangle is the distance from the midpoint to the point (0,0,4)

h = sqrt((2/3-0)2+(2/3-0)2+(0-4)2) = 4.1096093 (no real way to simplify)

the area of the triangle is

(0.5)base*height = 0.5(1.8856181)*(4.1096093) = 3.8745768

the flux through the triangle (plane) is

flux per unit area * area = 18/sqrt(19) * 3.8745768 = 16

the flux = 16

13. Suppose F is a radial force field, S1 is a sphere of
radius 9 centered at the origin, and the flux integral
Double_Integral_S1 F·dS= 2.
Let S2 be a sphere of radius 36 centered at the origin, and consider
the flux integral
Double_integral_S2 F·dS.

(A) If the magnitude of F is inversely proportional to the
square of the distance from the origin,what is the value of
Double_Integral_S2 F·dS?

well this one is easy!

Surface area = 4*pi*R2 so is proportional to R2

If F oc 1/R2

(oc is short hand for "proportional to")

then the surface integral is oc R2/R2 = constant

surface integral is a constant independent of R

S1 = S2 = 2

(B) If the magnitude of F is inversely proportional to the cube
of the distance from the origin, what is the value of
Double_integral_S2 F·dS?

Here S is oc to R2

but F oc 1/R3

The integral S2 oc R2/R3 = 1/R

If R increases by a factor of 36/9 = 4

then the surface integral decreases by the factor 1/4

S2 = S1/4 = 2/4 = 1/2

 

Expert:  MrBill31 replied 6 years ago.

11 confuses me since it is not a closed surface

will work on the rest and then return to it

 

12. Let S be the part of the plane 3x+3y+z = 4 which
lies in the first octant, oriented upward. Find the flux of the
vector field F = 2i+3j+3k across the surface S.

So we have a plane with a unit normal

(1/sqrt(32+32+12)(3i + 3j + k) = (1/sqrt(19))(3i + 3j + k)

it is bounded by

3x+3y+z = 4

x>=0

y>=0

z>=0

which means it is a triangle bounded by the points

(4/3,0,0)

(0,4/3,0)

(0,0,4)

What is the flux per unit area?

FA = n*F = (1/sqrt(19))(3i + 3j + k)*(2i+3j+3k )

= (1/sqrt(19))(3*2+3*3+1*3) = 18/sqrt(19)

what is the area of the triangle?

The base of the triangle is given as the distance between the points

(4/3,0,0) and (0,4/3,0)

b = sqrt(2)*(4/3) = 1.8856181

The mid point on the base is:

(2/3,2/3,0) (average of the points on the base (4/3,0,0) and (0,4/3,0))

the height of the triangle is the distance from the midpoint to the point (0,0,4)

h = sqrt((2/3-0)2+(2/3-0)2+(0-4)2) = 4.1096093 (no real way to simplify)

the area of the triangle is

(0.5)base*height = 0.5(1.8856181)*(4.1096093) = 3.8745768

the flux through the triangle (plane) is

flux per unit area * area = 18/sqrt(19) * 3.8745768 = 16

the flux = 16

13. Suppose F is a radial force field, S1 is a sphere of
radius 9 centered at the origin, and the flux integral
Double_Integral_S1 F·dS= 2.
Let S2 be a sphere of radius 36 centered at the origin, and consider
the flux integral
Double_integral_S2 F·dS.

(A) If the magnitude of F is inversely proportional to the
square of the distance from the origin,what is the value of
Double_Integral_S2 F·dS?

well this one is easy!

Surface area = 4*pi*R2 so is proportional to R2

If F oc 1/R2

(oc is short hand for "proportional to")

then the surface integral is oc R2/R2 = constant

surface integral is a constant independent of R

S1 = S2 = 2

(B) If the magnitude of F is inversely proportional to the cube
of the distance from the origin, what is the value of
Double_integral_S2 F·dS?

Here S is oc to R2

but F oc 1/R3

The integral S2 oc R2/R3 = 1/R

If R increases by a factor of 36/9 = 4

then the surface integral decreases by the factor 1/4

S2 = S1/4 = 2/4 = 1/2

Problem 3 of the last homework is the correct answer

However I am not comfortable with the derivation.

I think you would be more interested in me spending time figuring out problem 11 in this set.

So I will think about problem 11

turns out it is not too hard

it is a closed surface

problem with me is I have to see what something looks like before I can solve it

.

 

Expert:  MrBill31 replied 6 years ago.

11 SEE AFTER PROBLEM 12

12. Let S be the part of the plane 3x+3y+z = 4 which
lies in the first octant, oriented upward. Find the flux of the
vector field F = 2i+3j+3k across the surface S.

So we have a plane with a unit normal

(1/sqrt(32+32+12)(3i + 3j + k) = (1/sqrt(19))(3i + 3j + k)

it is bounded by

3x+3y+z = 4

x>=0

y>=0

z>=0

which means it is a triangle bounded by the points

(4/3,0,0)

(0,4/3,0)

(0,0,4)

What is the flux per unit area?

FA = n*F = (1/sqrt(19))(3i + 3j + k)*(2i+3j+3k )

= (1/sqrt(19))(3*2+3*3+1*3) = 18/sqrt(19)

what is the area of the triangle?

The base of the triangle is given as the distance between the points

(4/3,0,0) and (0,4/3,0)

b = sqrt(2)*(4/3) = 1.8856181

The mid point on the base is:

(2/3,2/3,0) (average of the points on the base (4/3,0,0) and (0,4/3,0))

the height of the triangle is the distance from the midpoint to the point (0,0,4)

h = sqrt((2/3-0)2+(2/3-0)2+(0-4)2) = 4.1096093 (no real way to simplify)

the area of the triangle is

(0.5)base*height = 0.5(1.8856181)*(4.1096093) = 3.8745768

the flux through the triangle (plane) is

flux per unit area * area = 18/sqrt(19) * 3.8745768 = 16

the flux = 16

13. Suppose F is a radial force field, S1 is a sphere of
radius 9 centered at the origin, and the flux integral
Double_Integral_S1 F·dS= 2.
Let S2 be a sphere of radius 36 centered at the origin, and consider
the flux integral
Double_integral_S2 F·dS.

(A) If the magnitude of F is inversely proportional to the
square of the distance from the origin,what is the value of
Double_Integral_S2 F·dS?

well this one is easy!

Surface area = 4*pi*R2 so is proportional to R2

If F oc 1/R2

(oc is short hand for "proportional to")

then the surface integral is oc R2/R2 = constant

surface integral is a constant independent of R

S1 = S2 = 2

(B) If the magnitude of F is inversely proportional to the cube
of the distance from the origin, what is the value of
Double_integral_S2 F·dS?

Here S is oc to R2

but F oc 1/R3

The integral S2 oc R2/R3 = 1/R

If R increases by a factor of 36/9 = 4

then the surface integral decreases by the factor 1/4

S2 = S1/4 = 2/4 = 1/2

Problem 3 of the last homework is the correct answer

However I am not comfortable with the derivation.

I think you would be more interested in me spending time figuring out problem 11 in this set.

So I will think about problem 11

turns out it is not too hard

it is a closed surface

problem with me is I have to see what something looks like before I can solve it

11. Evaluate Double_Integral_S sqrt(1+x^2+y^2) dS where S is the helicoid:
r(u,v) = ucos(v)i+usin(v)j+vk, with 0 <= u <= 1,0 <= v <= 3pi

this one works best in cylindrical coordinates

The thing is bounded by r = 0 to 1

z = 0 to 3pi

and makes 1.5 turns as it spirals upward

note the angle it makes with the xy plane is 45 degrees

(for each amount dz it moves up it has an arc length rdθ = dz)

so each unit of surface area has a factor of sqrt(2) compared to a circle on the xy plane

so in terms of r and θ the surface area element is

sqrt(2)rdrdθ

we are integrating sqrt(1+x2+y2) = sqrt(1+r2)

so the integral becomes

Int(r=0 to 1, θ = 0 to 3 π)sqrt(1+r2)*sqrt(2)rdrdθ]

=sqrt(2) Int[(r=0 to 1, θ = 0 to 3 π)rsqrt(1+r2)*drdθ]

=sqrt(2) 3 π Int[(r=0 to 1)rsqrt(1+r2)*dr]

= sqrt(2)3 π (1/3)(1+r2)3/2(r=1 - r=0)

= sqrt(2)π(23/2-1) = π(4-sqrt(2))

please press "accept" to accept this answer

thanks

mrbill

 

Customer: replied 6 years ago.
the answer for number 11 is wrong.

I assume that n represent pi right? if so the answer is wrong.

I don't know maybe this equation might help. this is what my teacher gave us today.

Double_integral_S (F dS) = Double_Integral_D (F(r(u,v))abs(ru x rv)) dA
Expert:  MrBill31 replied 6 years ago.
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