# 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!

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