# Using divergence theorem?

1. Apr 25, 2015

### uzman1243

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

2. Relevant equations

3. The attempt at a solution
I thought of using the divergence theorem where

I find that ∇.F = 3z
thus integral is
∫ ∫ ∫ 3z r dz dr dθ where r dz dr dθ is the cylindrical coordinates
with limits
0<=z<=4
0<=r<=3
0<=θ<=2π

and solving gives me 216π

Can I use this method for this question? I'm skeptical about my method because the question talks about normal unit vector and paramatizing a surface

2. Apr 25, 2015

### BvU

Well, this is a nice opportunity to verify the divergence theorem !
In principle you're free to ignore the hint, but I'm afraid some bad-tempered graders might think otherwise.

If you want to explore the path hinted at: what would you write for $dS$ at $(x,y,z) = (r \cos \phi, r\sin \phi, z)$

(sorry for my genetically fixed notation for azimuthal angles )

3. Apr 25, 2015

### uzman1243

So my method is valid?

so $dS$ = N dr d ∅

4. Apr 25, 2015

### Zondrina

By the divergence theorem:

$$\iint_S \vec F \cdot d \vec S = \iiint_V \text{div}(\vec F) \space dV = \iiint_V 3z \space dV$$

The solid volume $V$ is the cylinder that sweeps the angle $0 \leq \theta \leq 2 \pi$, has radius 3, and height 4. So yes the answer would be $216 \pi$.

The question wants you to do this problem a little bit differently though. You need to parametrize the cylinder like so:

$$\vec r(\theta, z) = 3\text{cos}(\theta) \hat i + 3\text{sin}(\theta) \hat j + z \hat k$$

Then using this theorem:

$$\iint_S \vec F \cdot d \vec S = \iint_D \vec F(\vec r(\theta, z)) \cdot (\vec r_{\theta} \times \vec r_z) \space dA$$

The integral can be evaluated to obtain the same result by using polar co-ordinates.

5. Apr 25, 2015

### pasmith

Technically the divergence theorem is not available here, since the surface S is not a closed surface.

6. Apr 25, 2015

### BvU

We were going to find out by evaluating the alternative route.
However, the comment by pasmith predicts that the answers differ.​

N being North ?

My plan was to do the inner product in Cartesian coordinates (I'm rather lazy). But Z's theorem is also intriguing (where does it come from ?).

--

7. Apr 25, 2015

### Zondrina

If $S$ is a smooth orientable surface given in parametric form by a vector function $\vec r(u,v)$, then the orientation induced by $\vec r(u,v)$ is defined by the unit normal vector;

$$\vec n = \frac{\vec r_u \times \vec r_v}{|\vec r_u \times \vec r_v|}$$

This vector is oriented out of the surface.

Now, if $\vec F(x,y,z)$ is a continuous vector field defined on the oriented surface $S$ with normal vector $\vec n$, we can re-write the flux integral:

$$\iint_S \vec F(x,y,z) \cdot d \vec S = \iint_S \vec F(x,y,z) \cdot \vec n \space dS = \iint_D \vec F(\vec r(u, v)) \cdot \frac{\vec r_u \times \vec r_v}{|\vec r_u \times \vec r_v|} |\vec r_u \times \vec r_v| \space dA = \iint_D \vec F(\vec r(u, v)) \cdot (\vec r_{u} \times \vec r_v) \space dA$$

So the flux integral of the vector field across the surface $S$ is equivalent to the double integral of the parametrized vector field over the region $D$.

Also, I believe the surface is closed because it is the boundary of a solid region. Here is the surface:

The flux out of the top and bottom surfaces will cancel due to the orientations.

Last edited: Apr 25, 2015
8. Apr 25, 2015

### Staff: Mentor

I don't think so. The vector field is different on the top surface than the bottom surface, particularly the z component, which is what is dotted with the normal.

If I were doing this problem, I would use the divergence theorem to allow me to integrate over the entire volume, and then I would subtract the net flux through the top and bottom surfaces. The latter would come out to 144π (basically out the top surface, since the flux out the bottom surface is zero).

Chet

9. Apr 26, 2015

### uzman1243

Ok so im trying the surface integral
r(θ,z) = 3cosθ i + 3sinθ j + zk

dr/dθ = -3sinθ i + 3cosθ j +0 k
dr/dz = 0i + 0j + 1k
dr/dθ x dr/dz = 3cosθ, 3sinθ,0

|dr/dθ x dr/dz| = 3

thus I get $$\vec n$$ = cosθ,sin θ,0

finally adding into the integral
∫∫ (3zcosθ,3cosθ,z2).(cosθ,sinθ,0) .3 dθ dz where 0<=z<=4 and 0<=θ<=2pi and 3 being the jacobian

and then finally solving gives me 72pi

Im guessing I've gone wrong somewhere?

Last edited: Apr 26, 2015
10. Apr 26, 2015

### uzman1243

I cannot understand what you mean. Can you please explain a little more?

11. Apr 26, 2015

### BvU

You worked out the $\hat n$ vector all right, but the dS is not $d\phi dz$ (or $d\theta dz$ as mathematicians insist on calling it ).
imho this $\hat n$ vector is pretty obvious also when not using Z's theorem
Where you go wrong in working out the integral, I can't really see because of my limited telepathic powers , so if you want guidance there, you should post your working in more detail. $27 \pi$ is definitely not what I got (and I did get a lot of different answers trying to check the 216$\pi$ at first, because I totally missed pasmith's insight!). Mathematically I wonder from where you got two extra factors 3

And the wonder of it all is that your initial answer and Chet's correction for the end surfaces match the outcome (well, at least they match what I found).

Economically speaking, the divergence theorem isn't the optimal way: you have to evaluate two integrals thanks to the obstacle pasmith pointed out. Going straight only requires one integral...

12. Apr 26, 2015

### uzman1243

Sorry I made some changes a few minutes ago. Can you check my reply above again?

13. Apr 26, 2015

### BvU

Yeah, now the thread has become labyrinthic. You inserted the r = 3 and oops, 27 changes in 72. Even though 3 x 27 is not 72.

My telepathic powers are sufficient to claim that
• in your workings you did not forget the factor r
• and the 27 was just a typo.
So even without having seen the detailed steps I wager you did just fine. (Jacobian ? I just call $rd\phi dz$ by the name $dS$ and imagine a curved little rectangular window ... goes to show how primitively physicists think)

(I type with 2 x 1.5 fingers and still one hand is often too quick for the other one) .

So all's good that ends good. That 216 = 144 + 72 Is a nice reassurance in this exercise ! Well done!

14. Apr 26, 2015

### uzman1243

where are you getting the 144 from?

15. Apr 26, 2015

### BvU

A most reliable source: post #8. And it's not a difficult calculation, so I did it too.

16. Apr 26, 2015

### uzman1243

I don't get it.
using divergence theorem I get 216pi. Using the surface integrals I get 72pi. The answer to solving using surface integrals should give me the same answer as using divergence theorem (provided it's a close surface).

17. Apr 26, 2015

### BvU

Using the divergence theorem you get 216$\;\pi$. The top and bottom surface give 144$\;\pi$. The exercise wants only the side surface (although it doesn't state that all too explicitly, I must admit).

Reading this problem statement once again, I realize that interpretation is forced upon me by my own context: cylindrical membranes where you pump something in and something goes out through the membrane; the rest goes out at the other end. A cell biologist would have a totally different picture, I imagine. So the mathematician has to make sure the problem statement is unambiguous.

Last edited: Apr 26, 2015
18. Apr 26, 2015

### uzman1243

so the answer is just 72 pi?

19. Apr 26, 2015

### BvU

What I'm trying to say is: I don't know. The most literal mathematical interpretation of the exercise text would favor the 216.

But then I don't really understand the hint in the problem statement, because in that case the divergence route you took is more economical.

20. Apr 26, 2015

### Delta²

No i think the exercise text means only the side surface cause it says $x^2+y^2=9$.(notice the equality, its not <=) The points with (x,y) such that satisfy that equation lie all in the side surface, otherwise it would have state something like to include the bottom and up surfaces with $x^2+y^2\leq 9,z=0,z=4$.

So the answer should be just 72pi.

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