Using the Divergence Theorem to Find Flux

  • Thread starter Raen
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  • #1
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Let W be the solid bounded by the paraboloid x = y^2 + z^2 and the plane x = 16. Let = 3xi + yj + zk
a. Let S1 be the paraboloid surface oriented in the negative x direction. Find the flux of the vector field through the surface S1.

b. Let S be the closed boundary of W. Use the Divergence Theorem to find the total flux out of S.


divF = dF1/dx + dF2/dy + dF3/dz
fluxF = Int(FdA) = Int(divFdV) (Sorry, the integral tags aren't working for me.)

I found the divF easily enough to be 5, but when I try to use that to calculate the flux, the online homework says that I'm wrong.

Int(divFdV) should be the integral of 5 over the volume of the paraboloid, which is x from 0 to 16, y from -4 to 4, and z from -4 to 4. Integrating over x gives 80, then over y gives 640, then over z gives 5120, which should be the answer. I tried -5120 as well, since it says the surface is oriented in the negative x direction, but that is also wrong. Can anybody tell me what I'm messing up? Also, once I find the flux, that should be the answer for both questions, right?

Sorry for the somewhat messy explanation of my work, I'm not used to explaining my process without being able to draw things out.
 

Answers and Replies

  • #2
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I would re-check the limits used on the integral:
If you think (graphically) of the meaning of these limits, you will see that it's indeed a 'rectangle' with base 16, depth 8 and height 8. And that's definitely not your part of the paraboloid.

I would suggest you to play around with the given parameters (x = y^2 + z^2 and the plane x = 16), so that in the first two limits your 'boundary' depends on variable(s); and then, in the last limit, it should be x from 0 to 16 (or equivalent s.t. the integral gives a fixed result). I think it would be easier to integrate using cylindrical coordinates too. But that's completely up to you.
 
  • #3
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How do I use cylindrical coordinates if it's not vertical, though? That was my first thought, but since the equation is x = y^2 + z^2 instead of z = x^2 + y^2 I wasn't sure how to write it out.
 
  • #4
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But don't you think it should be the same, just with a change of base?

I mean, using the 'normal' definition, it is:
z=z
x=rcosφ
y=rsinφ

with jacobian = r

Note that in our case we would rather have: x = r^2 = y^2 + z^2
Then, a good guess would be (according to symmetry in the x-z axis):

x=x
y=rsinφ
z=rcosφ

Now, you should get the limits with this. I mean, if you can imagine the paraboloid itself, things get rather easy. You can see that 0<x<r^2 for dx. Then, from this, we get: 0<r<4 for dr (since x was going from 0 to 16).

Also, a) and b) should give the same result: true (even though S1 is oriented negative, so maybe there will be some sign differences); but that doesn't mean that by 'just' using the RHS of the divergence theorem you are done. This means that you have done b). But for a), I guess that they want you to calculate the double integral. (i.e. the LHS of the divergence theorem– and more lengthy.)


P.S. Oh, now that I noticed: welcome to PhysicsForums ;)
 
  • #5
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I didn't mention the dφ in the last post, but I'm sure you have taken care of. If you want to integrate through the whole range of dφ, just go with the good old 0<φ<2π boundary.
 
  • #6
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So using cylindrical coordinates, I integrate 5r over r, theta, and x. The integral over r gives 40, the integral over theta gives 80pi, and the integral over x gives 1280pi, which would be my final answer?

And thank you for the welcome. :)
 
  • #7
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What do you mean by which should be your final answer?

When you do a triple integral you have:

[tex]\int^{2\pi}_{0} \int^{4}_{0} \int^{r^2}_0 5r dx dr d\phi[/tex]

then, this gives you (doing the first integral):

[tex]\int^{2\pi}_{0} \int^{4}_{0} 5r (r^2) dr d\phi [/tex]

etc.

I think you have got something wrong in the concept of triple integral. When you do a triple integral, you are just integrating something three times. So it should give you just one final result. Let me know if you have more questions concerning triple integrals.
 
  • #8
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I wasn't asking which as in "which one?" I was stating the final answer from the triple integral, and checking that it should be the final answer for the problem. I wanted to be sure there wasn't a step after the integral I was missing. However, the answer I got (1280pi) is incorrect. The negative is also correct, so either something is wrong with my math (although I used a graphing calculator to check), something is wrong with my limits (although you used the same numbers), or the way to find the correct answer.
 
  • #9
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Hm, I am getting 640/3 pi (provided that the divergence is correct).

Which steps are you doing to integrate?
 

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