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Divergence Theorem/Flux Integral Help!

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute the flux of F=xi+yj+zk through the curved surface of the cylinder x2+y2=1 bounded below by the plane x+y+z=1, above by the plane x+y+z=7, and oriented away from the z-axis.


    2. Relevant equations
    div(F) = (dF/dx) + (dF/dy) + (dF/dz)



    3. The attempt at a solution
    Basically all I did was set up a triple integral after finding div(F=3.

    ∫∫∫ (3) dz dy dx

    Limits of the right most integral = 1-x-y to 7-x-y
    Limits of the middle integral: -sqrt(1-x^2) to sqrt(1-x^2)
    Limits of the left integral: -1 to 1
    I got 18(pi) as my answer but it's wrong. Any help on this?
     
  2. jcsd
  3. Jun 15, 2012 #2
    You would have to show your work. It looks like your set up is OK. But I would advise switching to integrate in cylindrical coordinates instead. I can practically do it in my head, the height is 6 everywhere, over the area is π. So I think the answer is 18π also. Why do you think this is not the right answer?
     
  4. Jun 15, 2012 #3
    I also got 18pi, but since it's an online homework question, it says that I'm wrong. I have no idea why. Got 18pi in both Cartesian coordinates and cylindrical.
     
  5. Jun 15, 2012 #4

    vela

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    When you integrate the divergence, you're calculating the flux through all surfaces of the volume. The problem asked you to find the flux only through the curved surface of the cylinder.
     
  6. Jun 15, 2012 #5
    Does that mean I have to substract the flux from the bottom of the cylinder as well as the top? How would I go about doing that since they are planes intersecting the cylinder, and in essence I think are ellipses.
     
  7. Jun 15, 2012 #6
    I failed to read the question carefully, thank you vela!

    That surface integral does not look nice just looking at it, may want to try

    [tex]\iint_A+\iint_B=\iint_{A\cup B}=\iiint_V[/tex]

    Where the triple integral is 18π as we found.

    [itex]A[/itex] is the round part of the surface, which may be at least annoying to compute,

    [itex]B[/itex] is the two planar lids to the volume, which looks easier to calculate.
     
  8. Jun 15, 2012 #7
    Their shadows are not ellipses.
     
  9. Jun 15, 2012 #8
    Oh okay. So basically parameterize 2 circles in terms of r and theta, where r goes from 0 to 1 and theta goes from 0 to 2*pi ? And do the integration from there?
     
  10. Jun 15, 2012 #9
    Alright, thanks guys, got the question correct.

    Just did a double integral, with the integrand 3r dr d(theta) with the limits for dr going from 0 to 1 and for theta, 0 to 2*pi. Multiplied that by 2 since we have a top and bottom to take care of, and subtracted that (which was 6*pi) from 18*pi.
     
  11. Jun 15, 2012 #10
    [STRIKE]Now that I think about it, I'm a little worried, what is [itex]\textbf{F}\cdot d\textbf{A}[/itex] along the lids?[/STRIKE]

    was repeated below
     
  12. Jun 15, 2012 #11
    Now that I think about it, I'm a little worried, what is [itex]\textbf{F}\cdot d\textbf{A}[/itex] along the lids? But if you notice some niceties, the x and y contributions will cancel, and you just need the difference in z again, which is 6, so 6π. I'm not sure you used the right steps to get the right answer, it appears you integrate the divergence over the lids, that doesn't seem right.
     
  13. Jun 15, 2012 #12
    Really? I have no idea how that happened, but that's one heck of a coincidence....

    did I have to do the integrated of F*dA along the lids to get the answer?
     
  14. Jun 15, 2012 #13
    Yes. So typically you should have a formula for [itex]d\textbf{A}[/itex] when you parametrize the surface, which shouldn't be hard, or you can look at the surface as a "graph" of z(x,y) (special case of the more general paramtrization), and they have a neat formula for that.
     
  15. Jun 15, 2012 #14
    So even in that case I'd be just looking at the shadows of the elliptical lids (aka circles)?
     
  16. Jun 15, 2012 #15
    yes, if you're parametrization or graph was over x-y plane. Rememeber that is due to an absolute constraint by the circular cylinder.
     
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