Surface integral (Flux) with cylinder and plane intersections

[CAF123]

Homework Statement

Evaluate the surface integral \int_{S} \int \vec{F} \cdot \vec{n}\,dSwith the vector field \vec{F⃗}=zx\vec{i}+xy\vec{j}+yz\vec{k}. S is the closed surface composed of a portion of the cylinderx^2 + y^2 = R^2that lies in the first octant, and portions of the planes x=0, y=0, z=0 and z=H. \vec{n} 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: <br /> \int_{0}^{R} \int_{0}^{\sqrt{R^2- x^2}} -yz\, dy\,dx
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 \int_{0}^{R} \int_{0}^{H} -xy\,dz\,dx 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.

 

[Dick]

Homework Statement

Evaluate the surface integral \int_{S} \int \vec{F} \cdot \vec{n}\,dSwith the vector field \vec{F⃗}=zx\vec{i}+xy\vec{j}+yz\vec{k}. S is the closed surface composed of a portion of the cylinderx^2 + y^2 = R^2that lies in the first octant, and portions of the planes x=0, y=0, z=0 and z=H. \vec{n} 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: <br /> \int_{0}^{R} \int_{0}^{\sqrt{R^2- x^2}} -yz\, dy\,dx
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 \int_{0}^{R} \int_{0}^{H} -xy\,dz\,dx 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.

 

[CAF123]

Hi Dick,
So this means the surface integral over S2,S3,S4 would all be zero?
For S5, I have \int_{0}^{R} \int_{0}^{\sqrt{R^2-x^2}}\,yz\,dy\,dx, which when evaluated gives 1/3 HR^3 since z=H.
For S1, the cylinder part, I used cylindrical coords. I used the parametrisation \vec{r}(R,\theta) = R\cos\theta \vec{i} + R\sin\theta \vec{j} + \vec{k}. This gives r_{R} = \cos\theta \vec{i} + \sin\theta\vec{j} and r_{\theta} = -R\sin\theta \vec{i} + R\cos\theta \vec{j}. Taking these as vectors in R3 with z component 0 gives a cross product of R\vec{k}. Dotting this with the vector field given parametrized in terms of R and theta give F \cdot (r_R × r_{\theta}) = R^2 \sin\theta z. This sets up :\int_{0}^{\frac{\pi}{2}} \int_{0}^{R} \int_{0}^{H} R^2 \sin\theta z\,R\,dz\,dR\,d\theta, 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?

 

[Dick]

Hi Dick,
So this means the surface integral over S2,S3,S4 would all be zero?
For S5, I have \int_{0}^{R} \int_{0}^{\sqrt{R^2-x^2}}\,yz\,dy\,dx, which when evaluated gives 1/3 HR^3 since z=H.
For S1, the cylinder part, I used cylindrical coords. I used the parametrisation \vec{r}(R,\theta) = R\cos\theta \vec{i} + R\sin\theta \vec{j} + \vec{k}. This gives r_{R} = \cos\theta \vec{i} + \sin\theta\vec{j} and r_{\theta} = -R\sin\theta \vec{i} + R\cos\theta \vec{j}. Taking these as vectors in R3 with z component 0 gives a cross product of R\vec{k}. Dotting this with the vector field given parametrized in terms of R and theta give F \cdot (r_R × r_{\theta}) = R^2 \sin\theta z. This sets up :\int_{0}^{\frac{\pi}{2}} \int_{0}^{R} \int_{0}^{H} R^2 \sin\theta z\,R\,dz\,dR\,d\theta, 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?

 

[CAF123]

My book has the general method to find the surface integral: \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. 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 | r_{R} × r_{\theta} | cancelled

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

 

[Dick]

My book has the general method to find the surface integral: \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. 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 | r_{R} × r_{\theta} | 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.

 

[CAF123]

So I had the parametrisation wrong? Should it have been r (θ, z) = Rcosθi + Rsinθj + zk?

 

[Dick]

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?

 

[CAF123]

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?

 

[Dick]

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.

 

[CAF123]

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 \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 is this ok? Why are the limits dz and Rdθ rather than just dz and dθ?

 

[Dick]

I have \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 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.

 

[CAF123]

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)

 

[Dick]

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?

 

[CAF123]

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 R^2 \cos^2 \theta z + R^3\cos\theta\sin^2\theta Rdzdθ

 

[Dick]

I made a mistake when quoting what I had to integrate. It should have been the double integral of R^2 \cos^2 \theta z + R^3\cos\theta\sin^2\theta Rdzdθ

Then the second term in your integrand picked up extra R someplace.

 

[Dick]

My book has the general method to find the surface integral: \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. 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 | r_{R} × r_{\theta} | 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!

 

[CAF123]

Then the second term in your integrand picked up extra R someplace.

What I said was 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

Edit: thanks for clarifying things. Ok, I will check my work

 

[Dick]

What I said was 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

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.

 

[Dick]

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!

 

[CAF123]

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.

 

[Dick]

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.

 

[CAF123]

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.

 

[Dick]

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.

\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.

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.

 

[CAF123]

\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.

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.

 

[Dick]

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.

 

[CAF123]

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

 

[Dick]

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.

 

[CAF123]

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.

 

[Dick]

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.

 

[CAF123]

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 $dA$?

 

[Dick]

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.

 

[CAF123]

Did you catch my edit in the last post?

 

[Dick]

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 $r dr d\theta$.

 

[CAF123]

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?

 

[Dick]

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?

They are two different uses of the symbol 'dA'. They are two different things.

 

[CAF123]

They are two different uses of the symbol 'dA'. They are two different things.

Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?

 

[Dick]

Ok, thanks. I think this also answers one of my other questions: The form dA=dr dθ is not really an area element because the dimensions don't check out. What is the difference between the two forms?

We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.

 

[CAF123]

We've been over that several times already. dr dθ is the area element without the jacobian part 'r'. You use that in the cross product form of the surface integral because the cross product contains the jacobian part, 'r'. If you are just integrating a function in polar coordinates you want the full area element, i.e. with jacobian r dr dθ.

Ok, everything makes sense. Thanks so much!