Surface integral (Flux) with cylinder and plane intersections

In summary, the conversation discusses evaluating a surface integral with a given vector field over a closed surface composed of a portion of a cylinder and portions of planes. The attempt at a solution includes setting up integrals for each surface and using cylindrical coordinates. However, there are some mistakes in the normal vectors and the results do not make sense. Further guidance is requested on how to tackle the remaining surfaces.
  • #1
CAF123
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Homework Statement



Evaluate the surface integral [tex] \int_{S} \int \vec{F} \cdot \vec{n}\,dS[/tex]with the vector field [itex] \vec{F⃗}=zx\vec{i}+xy\vec{j}+yz\vec{k} [/itex]. S is the closed surface composed of a portion of the cylinder[itex] x^2 + y^2 = R^2 [/itex]that lies in the first octant, and portions of the planes x=0, y=0, z=0 and z=H. [itex] \vec{n} [/itex] is the outward unit normal vector.
2.Attempt at a solution

Attempt: I said S consisted of the five surfaces S1,S2,S3,S4 and S5. S1 being the portion of the cylinder, S2 being where the plane z=0 cuts the cylinder and similarly, S3,S4,S5 being where the planes x=0,y=0 and z=H cut the cylinder.

For S2, the normal vector points in the -k direction. so the required integral over S2 is: [tex]
\int_{0}^{R} \int_{0}^{\sqrt{R^2- x^2}} -yz\, dy\,dx [/tex]
Am I correct? I think for the surface S5 the only thing that would change in the above would be that the unit normal vector points in the positive k direction?

I need some guidance on how to set up the integrals for the rest of the surfaces.(excluding the cylinder part - I used cylindrical coords here and have an answer) I tried [itex] \int_{0}^{R} \int_{0}^{H} -xy\,dz\,dx [/itex] for the y = 0 plane intersection with the cylinder, but I am not sure if this is correct.

Any advice on how to tackle the remaining surfaces would be very helpful.
 
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  • #2
CAF123 said:

Homework Statement



Evaluate the surface integral [tex] \int_{S} \int \vec{F} \cdot \vec{n}\,dS[/tex]with the vector field [itex] \vec{F⃗}=zx\vec{i}+xy\vec{j}+yz\vec{k} [/itex]. S is the closed surface composed of a portion of the cylinder[itex] x^2 + y^2 = R^2 [/itex]that lies in the first octant, and portions of the planes x=0, y=0, z=0 and z=H. [itex] \vec{n} [/itex] is the outward unit normal vector.
2.Attempt at a solution

Attempt: I said S consisted of the five surfaces S1,S2,S3,S4 and S5. S1 being the portion of the cylinder, S2 being where the plane z=0 cuts the cylinder and similarly, S3,S4,S5 being where the planes x=0,y=0 and z=H cut the cylinder.

For S2, the normal vector points in the -k direction. so the required integral over S2 is: [tex]
\int_{0}^{R} \int_{0}^{\sqrt{R^2- x^2}} -yz\, dy\,dx [/tex]
Am I correct? I think for the surface S5 the only thing that would change in the above would be that the unit normal vector points in the positive k direction?

I need some guidance on how to set up the integrals for the rest of the surfaces.(excluding the cylinder part - I used cylindrical coords here and have an answer) I tried [itex] \int_{0}^{R} \int_{0}^{H} -xy\,dz\,dx [/itex] for the y = 0 plane intersection with the cylinder, but I am not sure if this is correct.

Any advice on how to tackle the remaining surfaces would be very helpful.

I think you doing fine so far in setting them up. But some of them don't take much work to evaluate. Put z=0 into the first integral for S2 and y=0 into the second.
 
  • #3
Hi Dick,
So this means the surface integral over S2,S3,S4 would all be zero?
For S5, I have [tex] \int_{0}^{R} \int_{0}^{\sqrt{R^2-x^2}}\,yz\,dy\,dx, [/tex] which when evaluated gives 1/3 HR^3 since z=H.
For S1, the cylinder part, I used cylindrical coords. I used the parametrisation [itex] \vec{r}(R,\theta) = R\cos\theta \vec{i} + R\sin\theta \vec{j} + \vec{k}. [/itex] This gives [itex] r_{R} = \cos\theta \vec{i} + \sin\theta\vec{j} [/itex] and [itex] r_{\theta} = -R\sin\theta \vec{i} + R\cos\theta \vec{j}.[/itex] Taking these as vectors in R3 with z component 0 gives a cross product of [itex] R\vec{k} [/itex]. Dotting this with the vector field given parametrized in terms of R and theta give [itex] F \cdot (r_R × r_{\theta}) = R^2 \sin\theta z [/itex]. This sets up :[tex] \int_{0}^{\frac{\pi}{2}} \int_{0}^{R} \int_{0}^{H} R^2 \sin\theta z\,R\,dz\,dR\,d\theta, [/tex] which when evaluated gives 1/8H^2 R^4.
Adding the two results together gives HR^3(1/3 + 1/8 HR). Can anyone confirm this is correct?
 
  • #4
CAF123 said:
Hi Dick,
So this means the surface integral over S2,S3,S4 would all be zero?
For S5, I have [tex] \int_{0}^{R} \int_{0}^{\sqrt{R^2-x^2}}\,yz\,dy\,dx, [/tex] which when evaluated gives 1/3 HR^3 since z=H.
For S1, the cylinder part, I used cylindrical coords. I used the parametrisation [itex] \vec{r}(R,\theta) = R\cos\theta \vec{i} + R\sin\theta \vec{j} + \vec{k}. [/itex] This gives [itex] r_{R} = \cos\theta \vec{i} + \sin\theta\vec{j} [/itex] and [itex] r_{\theta} = -R\sin\theta \vec{i} + R\cos\theta \vec{j}.[/itex] Taking these as vectors in R3 with z component 0 gives a cross product of [itex] R\vec{k} [/itex]. Dotting this with the vector field given parametrized in terms of R and theta give [itex] F \cdot (r_R × r_{\theta}) = R^2 \sin\theta z [/itex]. This sets up :[tex] \int_{0}^{\frac{\pi}{2}} \int_{0}^{R} \int_{0}^{H} R^2 \sin\theta z\,R\,dz\,dR\,d\theta, [/tex] which when evaluated gives 1/8H^2 R^4.
Adding the two results together gives HR^3(1/3 + 1/8 HR). Can anyone confirm this is correct?

The first one looks ok. For the second one you've got the normal wrong. It isn't k. (And don't forget you need UNIT normals.) Your r_R vector isn't a tangent to the surface. What is it?
 
  • #5
My book has the general method to find the surface integral: [tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex] I have followed this to get my answer so I don't see what I have done wrong.
However, I see that the results I got don't make sense. I think n should be pointing outwards from the cylinder with some x and y components.
Also, my r_R describes a point on the unit circle. I see why this is not tangent.
I realize that the normal vectors should be normalised but when I studied the proof of the result above, the factor of [itex] | r_{R} × r_{\theta} | [/itex] cancelled

So where did I go wrong in the method that I followed?
 
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  • #6
CAF123 said:
My book has the general method to find the surface integral: [tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex] I have followed this to get my answer so I don't see what I have done wrong.
However, I see that the results I got don't make sense. I think n should be pointing outwards from the cylinder with some x and y components.
Also, my r_R describes a point on the unit circle. I see why this is not tangent.
I realize that the normal vectors should be normalised but when I studied the proof of the result above, the factor of [itex] | r_{R} × r_{\theta} | [/itex] cancelled

So where did I go wrong in the method that I followed?

The coordinates parameterizing your surface are theta and z. Not theta and r. r is a constant, r=R.
 
  • #7
So I had the parametrisation wrong? Should it have been r (θ, z) = Rcosθi + Rsinθj + zk?
 
  • #8
CAF123 said:
So I had the parametrisation wrong? Should it have been r (θ, z) = Rcosθi + Rsinθj + zk?

Sure. That seems more correct to you as well, yes?
 
  • #9
Yes. I should have saw it before.
So I just do the same as I did before but this time using my new parametrisation? (have i set up the triple integral correctly that i had previously - in terms of limits and transformation, r dzdRdθ? ) Are you able to give me an answer that I can check?
 
  • #10
CAF123 said:
Yes. I should have saw it before.
So I just do the same as I did before but this time using my new parametrisation? (have i set up the triple integral correctly that i had previously - in terms of limits and transformation, r dzdRdθ? ) Are you able to give me an answer that I can check?

It's only going to be a double integral. No integral over R. Just dz and Rdθ. I think you know limits. If you tell me what you get, I'll tell you whether it's right.
 
  • #11
Dick said:
It's only going to be a double integral. No integral over R. Just dz and Rdθ. I think you know limits. If you tell me what you get, I'll tell you whether it's right.
I have [tex] \int_{0}^{\frac{\pi}{2}} \int_{0}^{H} R^2 \cos^2\theta z + R^2\cos\theta\sin^2\theta \,R \,dz\, d\theta [/tex] is this ok? Why are the limits dz and Rdθ rather than just dz and dθ?
 
  • #12
CAF123 said:
I have [tex] \int_{0}^{\frac{\pi}{2}} \int_{0}^{H} R^2 \cos^2\theta z + R^2\cos\theta\sin^2\theta \,R \,dz\, d\theta [/tex] is this ok? Why are the limits dz and Rdθ rather than just dz and dθ?

Looks good to me. If you think of making an angular displacement of dθ the actual distance you move is rdθ. dV=dr dz (rdθ). The 'r' part of dV belongs with the dθ.
 
  • #13
Dick said:
Looks good to me. If you think of making an angular displacement of dθ the actual distance you move is rdθ. dV=dr dz (rdθ). The 'r' part of dV belongs with the dθ.
Ah, I remember now. I get an answer of R^3 H(Hpi/8 + Rpi/6 + 1/3) in total (incorporating the other surface as well)
 
  • #14
CAF123 said:
Ah, I remember now. I get an answer of R^3 H(Hpi/8 + Rpi/6 + 1/3) in total (incorporating the other surface as well)

Not what I get. Your dimensions aren't coming out right. Every term should have a total of 4 R's and H's. I get 2HR^3/3+pi H^2 R^2/8. Could you check that?
 
  • #15
Dick said:
Not what I get. Your dimensions aren't coming out right. Every term should have a total of 4 R's and H's. I get 2HR^3/3+pi H^2 R^2/8. Could you check that?
I made a mistake when quoting what I had to integrate. It should have been the double integral of [itex] R^2 \cos^2 \theta z + R^3\cos\theta\sin^2\theta Rdzdθ [/itex]
 
  • #16
CAF123 said:
I made a mistake when quoting what I had to integrate. It should have been the double integral of [itex] R^2 \cos^2 \theta z + R^3\cos\theta\sin^2\theta Rdzdθ [/itex]

Then the second term in your integrand picked up extra R someplace.
 
  • #17
CAF123 said:
My book has the general method to find the surface integral: [tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex] I have followed this to get my answer so I don't see what I have done wrong.
However, I see that the results I got don't make sense. I think n should be pointing outwards from the cylinder with some x and y components.
Also, my r_R describes a point on the unit circle. I see why this is not tangent.
I realize that the normal vectors should be normalised but when I studied the proof of the result above, the factor of [itex] | r_{R} × r_{\theta} | [/itex] cancelled

So where did I go wrong in the method that I followed?

I see what's going on. dS=r dθ dz. You didn't normalize the unit vector so you are using the dA form above. dA=dθ dz. Drop the extra r in the integral. I was using the second form. Sorry for the confusion!
 
  • #18
Dick said:
Then the second term in your integrand picked up extra R someplace.
What I said was [tex] F(θ,z) \cdot (R\cos\theta i+ R\sin\theta j) = R\cos\theta z i + R^2\cos\theta \sin\theta j + R\sin\theta z k \cdot (R\cos\theta i + R\sin\theta j) = R^2 cos^2\theta + R^3 cos\theta sin^2 \theta[/tex]

Edit: thanks for clarifying things. Ok, I will check my work
 
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  • #19
CAF123 said:
What I said was [tex] F(θ,z) \cdot (R\cos\theta i+ R\sin\theta j) = R\cos\theta z i + R^2\cos\theta \sin\theta j + R\sin\theta z k \cdot (R\cos\theta i + R\sin\theta j) = R^2 cos^2\theta + R^3 cos\theta sin^2 \theta[/tex]

That's correct. See my last post. You don't need extra r in the volume element. I was using n=(cos(theta),sin(theta),0) the unit normal.
 
  • #20
CAF123 said:
Edit: thanks for clarifying things. Ok, I will check my work

Yours is still the wrong one. Use what you got but drop the extra R from the dS!
 
  • #21
Dick said:
That's correct. See my last post. You don't need extra r in the volume element. I was using n=(cos(theta),sin(theta),0) the unit normal.
Ok, I get your answer now. Thanks a lot for staying with me and for your time.
 
  • #22
CAF123 said:
Ok, I get your answer now. Thanks a lot for staying with me and for your time.

No problem. BTW checking dimensions on terms can be handy. R and H are both lengths. So all of your terms should have the same number of 'lengths' in them. So I could tell that your answer in post 13 MUST be wrong, even if I hadn't done the calculation.
 
  • #23
Just a quick question about one of the things here. I said ##dA = dz d\theta##. Why is this the case? In polars, we have ##dA = r\,dr\,d\theta## which makes sense as an area since ##\theta## is dimensionless.
 
  • #24
CAF123 said:
Just a quick question about one of the things here. I said ##dA = dz d\theta##. Why is this the case? In polars, we have ##dA = r\,dr\,d\theta## which makes sense as an area since ##\theta## is dimensionless.

[tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex]

If you look at the above the dS has the dimensions of area. In the cross product form the extra r is coming from the cross product.
 
  • #25
Dick said:
[tex] \int \int_{S} \vec{F} \cdot \vec{dS} = \int \int_{S} \vec{F} \cdot \vec{n} \,dS = \int \int_{D} \vec{F} \cdot (\vec{r_R} × \vec{r_\theta})\, dA. [/tex]

If you look at the above the dS has the dimensions of area. In the cross product form the extra r is coming from the cross product.
In general, is ##dS = ( r_u \times r_v)\,dA, ##where ##dA = du\,dv##. So in polars, I could say ##dS = r\,dr\,d\theta##, where ##dA= dr\,d\theta## if I am correct.
 
  • #26
CAF123 said:
In general, is ##dS = ( r_u \times r_v)\,dA, ##where ##dA = du\,dv##. So in polars, I could say ##dS = r\,dr\,d\theta##, where ##dA= dr\,d\theta## if I am correct.

That's it. Work out a simple problem like finding the area of the unit disk. You'll that that the cross product term is r.
 
  • #27
I was looking at some other site (referenced below) and they take dA=r dr dθ. Why is this?

http://calculus7.com/sitebuildercontent/sitebuilderfiles/polrcoordntesdblint.pdf
 
  • #28
CAF123 said:
I was looking at some other site (referenced below) and they take dA=r dr dθ. Why is this?

http://calculus7.com/sitebuildercontent/sitebuilderfiles/polrcoordntesdblint.pdf

Because they aren't using the form with the cross product in it.
 
  • #29
What's the other form you mentioned? When I thought about it later, I see that they were just converting from Cartesian to polar and so the r was the Jacobian.
 
  • #30
CAF123 said:
What's the other form you mentioned? When I thought about it later, I see that they were just converting from Cartesian to polar and so the r was the Jacobian.

Yes, r is the Jacobian. The cross product factor also represents the Jacobian.
 
  • #31
I see. So they are essentially using the form $$ dS = \frac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA = \hat{n} r\,dr\,d\theta$$
EDIT: if this is the case, then shouldn't ##|r_u\times r_v| dA = r\,dr\,d\theta, ##instead of just ##dA##?
 
Last edited:
  • #32
CAF123 said:
I see. So they are essentially using the form $$ dS = \frac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA = \hat{n} r\,dr\,d\theta$$

Yes, that's it.
 
  • #33
Did you catch my edit in the last post?
 
  • #34
CAF123 said:
I see. So they are essentially using the form $$ dS = \frac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA = \hat{n} r\,dr\,d\theta$$
EDIT: if this is the case, then shouldn't ##|r_u\times r_v| dA = r\,dr\,d\theta, ##instead of just ##dA##?

Assuming we are still on polar coordinates ## \vec{r}=(r cos \theta, r sin \theta, 0)## , ##|\vec{r}_r\times \vec{r}_\theta|=r##, ##dA= dr d\theta##. So ##|\vec{r}_r\times \vec{r}_\theta| dr d\theta## IS ##r dr d\theta##.
 
  • #35
Dick said:
Assuming we are still on polar coordinates ## \vec{r}=(r cos \theta, r sin \theta, 0)## , ##|\vec{r}_r\times \vec{r}_\theta|=r##, ##dA= dr d\theta##. So ##|\vec{r}_r\times \vec{r}_\theta| dr d\theta## IS ##r dr d\theta##.
I see that this implies that dS = r dr dθ. But in the example from the website they also have dA = r dr dθ. How does this follow if dA = dr dθ? How can it equal two different things?
 
<h2>1. What is a surface integral (flux) with cylinder and plane intersections?</h2><p>A surface integral with cylinder and plane intersections is a mathematical concept that calculates the flow of a vector field through a surface formed by the intersection of a cylinder and a plane. It is used in physics and engineering to determine the amount of fluid or energy passing through a specific surface.</p><h2>2. How is a surface integral (flux) with cylinder and plane intersections calculated?</h2><p>To calculate a surface integral with cylinder and plane intersections, you first need to find the unit normal vector to the surface. Then, you need to set up a double integral over the surface using the given vector field and the unit normal vector. Finally, you evaluate the integral using appropriate limits and integration techniques.</p><h2>3. What is the physical significance of a surface integral (flux) with cylinder and plane intersections?</h2><p>The physical significance of a surface integral with cylinder and plane intersections is that it represents the amount of fluid or energy passing through a specific surface. This can be useful in various applications, such as determining the flow of air over an airplane wing or the flow of water through a pipe.</p><h2>4. Can a surface integral (flux) with cylinder and plane intersections be negative?</h2><p>Yes, a surface integral with cylinder and plane intersections can be negative. This occurs when the vector field and the unit normal vector are in opposite directions, resulting in a negative value for the flux. This indicates that the flow is going in the opposite direction of the surface's orientation.</p><h2>5. What are some real-life examples where a surface integral (flux) with cylinder and plane intersections is used?</h2><p>A surface integral with cylinder and plane intersections is commonly used in fluid dynamics, such as in calculating the flow of air over an airplane wing or the flow of water through a pipe. It is also used in electromagnetism to calculate the electric or magnetic flux through a surface, and in heat transfer to determine the amount of heat passing through a surface.</p>

1. What is a surface integral (flux) with cylinder and plane intersections?

A surface integral with cylinder and plane intersections is a mathematical concept that calculates the flow of a vector field through a surface formed by the intersection of a cylinder and a plane. It is used in physics and engineering to determine the amount of fluid or energy passing through a specific surface.

2. How is a surface integral (flux) with cylinder and plane intersections calculated?

To calculate a surface integral with cylinder and plane intersections, you first need to find the unit normal vector to the surface. Then, you need to set up a double integral over the surface using the given vector field and the unit normal vector. Finally, you evaluate the integral using appropriate limits and integration techniques.

3. What is the physical significance of a surface integral (flux) with cylinder and plane intersections?

The physical significance of a surface integral with cylinder and plane intersections is that it represents the amount of fluid or energy passing through a specific surface. This can be useful in various applications, such as determining the flow of air over an airplane wing or the flow of water through a pipe.

4. Can a surface integral (flux) with cylinder and plane intersections be negative?

Yes, a surface integral with cylinder and plane intersections can be negative. This occurs when the vector field and the unit normal vector are in opposite directions, resulting in a negative value for the flux. This indicates that the flow is going in the opposite direction of the surface's orientation.

5. What are some real-life examples where a surface integral (flux) with cylinder and plane intersections is used?

A surface integral with cylinder and plane intersections is commonly used in fluid dynamics, such as in calculating the flow of air over an airplane wing or the flow of water through a pipe. It is also used in electromagnetism to calculate the electric or magnetic flux through a surface, and in heat transfer to determine the amount of heat passing through a surface.

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