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Flux and divergence theorem

  1. Apr 28, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img16.imageshack.us/img16/88/fluxm.th.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution

    I've tried to find the divergence of F and I got 3x^2 + 3y^2 + 3z^2 and as this is a variable I need to set up the integral... how do I set the integral
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 28, 2009 #2
    i think using cylindrical coordinate will be fine
     
  4. Apr 28, 2009 #3
    how would you do it in cylindrical coordinate? is it something like this?

    do you mean that I don't need to use the divF of it?
     
    Last edited: Apr 28, 2009
  5. Apr 28, 2009 #4
    you mean how to express the integrated region?just set r from 0 to 3, z from 0 to10.
    using divF is not a bad choice, indeed in this case it won't save you very much work,because it's pretty symmetric casem a you can try to calculate it directly,
     
  6. Apr 28, 2009 #5
    to use divergence I will need tot calculate the volume right? and I then will need to convert x, y, and z to cylindrical coordinate... which is x = r cos(theta), y = r cos(theta) , so in this case it's x^3cos(theta)^3.. right?
     
  7. Apr 28, 2009 #6

    djeitnstine

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    Rule of thumb...x axis and y axis are not parallel....y=r sin(theta)... unless you're dealing with some weird system...
     
  8. Apr 28, 2009 #7
    oops.. bad copy and paste mistake.. so in the end it will be:

    \int0^10 \int_0^3 27cos(\theta) + 27sin(\theta) + z^3 dr d\theta

    now what is r here?
     
  9. Apr 28, 2009 #8

    djeitnstine

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    This is a triple integral so there should be 3 limits. Also it seems that [tex]3x^2 + 3y^2 + 3z^2[/tex] is your field, so in cylindrical it would be [tex]\int_0^{2\pi} \int_0^3 \int_0^{10} 3r^2 cos^2(\theta) + 3r^2 sin^2(\theta) + 3z^2 r dz dr d\theta[/tex]

    As you can see the field cancels out nicely in cylindrical. Also note the extra 'r' and the order of the limits.

    The field seems symmetrical so you should be able to go from 0-->pi/2 and multiply by 2

    latex seems to be ok again
     
  10. Apr 28, 2009 #9
    so it could then be simplified as:

    [tex]\int_0^{2\pi} \int_0^3 \int_0^{10} 3r^2 + 3z^2 r dz dr d\theta[/tex]
    [tex]\int_0^{2\pi} \int_0^3 30r^2 + 1000r dr d\theta[/tex]
    [tex]\int_0^{2\pi} 4770 d\theta[/tex]
    [tex]9540\pi[/tex]

    is this wrong?

    my confusion is that why is it 3z^2r ?? why is there extra r at the end of z?
     
    Last edited: Apr 28, 2009
  11. Apr 28, 2009 #10

    djeitnstine

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    This is because when converting to spherical the area of the differential section is [tex]rdrd\theta[/tex] and adapted to cylindrical is [tex]rdzdrd\theta[/tex] since the differential z is linear
     
  12. Apr 28, 2009 #11
    the answer that I got above doesn't match.. wonder why..
     
  13. Apr 28, 2009 #12

    djeitnstine

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    Oh I needed brackets, I am sorry

    [tex]
    \int_0^{2\pi} \int_0^3 \int_0^{10} (3r^2 + 3z^2) r dz dr d\theta
    [/tex]

    r is ditributed
     
  14. Apr 28, 2009 #13
    that's why I ask first place why is it 3z^2r because usually r is distributed.. thanks for clearing that up
     
  15. Apr 28, 2009 #14
    here's what I got:

    [tex]\int_0^{2\pi} \int_0^3 \int_0^{10} (3r^3 + 3z^2r) dz dr d\theta [/tex]
    [tex]\int_0^{2\pi} \int_0^3 30r^3 + 27r dr d\theta [/tex]
    [tex]\int_0^{2\pi} 2916/4 d\theta [/tex]
    [tex]\frac{2916\pi}{2} [/tex]


    the answer still doesn't match.. am I doing something wrong here?
     
  16. Apr 28, 2009 #15
    thanks I got it now :)
     
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