Surface integral (Flux) with cylinder and plane intersections

1. Nov 3, 2012

CAF123

1. The problem statement, all variables and given/known data

Evaluate the surface integral $$\int_{S} \int \vec{F} \cdot \vec{n}\,dS$$with 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 cylinder$x^2 + y^2 = R^2$that 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: $$\int_{0}^{R} \int_{0}^{\sqrt{R^2- x^2}} -yz\, dy\,dx$$
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.

2. Nov 3, 2012

Dick

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. Nov 4, 2012

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?

4. Nov 4, 2012

Dick

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. Nov 4, 2012

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 realise 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?

Last edited: Nov 4, 2012
6. Nov 4, 2012

Dick

The coordinates parameterizing your surface are theta and z. Not theta and r. r is a constant, r=R.

7. Nov 4, 2012

CAF123

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

8. Nov 4, 2012

Dick

Sure. That seems more correct to you as well, yes?

9. Nov 4, 2012

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?

10. Nov 4, 2012

Dick

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. Nov 4, 2012

CAF123

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?

12. Nov 4, 2012

Dick

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. Nov 4, 2012

CAF123

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. Nov 4, 2012

Dick

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. Nov 4, 2012

CAF123

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

16. Nov 4, 2012

Dick

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

17. Nov 4, 2012

Dick

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. Nov 4, 2012

CAF123

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

Last edited: Nov 4, 2012
19. Nov 4, 2012

Dick

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. Nov 4, 2012

Dick

Yours is still the wrong one. Use what you got but drop the extra R from the dS!