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Is there any way to calculate this integral?

  1. May 24, 2017 #1
    • Moved from a technical forum, so homework template missing
    upload_2017-5-24_12-59-48.png
    I have done it by the parametric form of σ, but if I change σ to implicit form that is G(x,y,z)=x^2+y*2+z^2-R^2=0 I don't know how continue.
    The theory is:
    upload_2017-5-24_13-3-58.png
    where Rxy is the projection of σ in plane xy so it's the circumference x^2+y^2=R^2
     
  2. jcsd
  3. May 24, 2017 #2

    BvU

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    Hello Rafa, :welcome:

    Maybe the problem statement is correct, but you don't tell what you are actually doing (parametric form of ##\sigma## ?)
    Maybe the theory is correct, but you don't explain all the symbols, so to me it's of no practical use.
    In the past I learned to change to spherical coordinates for something like this and if I do that here I have no problem coming up with the answwer. How about you ?

    [edit] Oh, and: for homework you should post in the homework forum and use the template there.
     
  4. May 24, 2017 #3
    Thanks,
    I want to calculate this integer but σ:x^2+y^2+z^2-R^2=0
     
  5. May 24, 2017 #4

    ehild

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    You can do the integral in spherical polar coordinates. http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html
    Write x and y and also the surface element in the spherical coordinates.
     
  6. May 24, 2017 #5
    Can you write it please?
     
  7. May 24, 2017 #6

    BvU

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    PF requires you to do something too. You said you had done it
    Show your work

    [edit] on a friendlier note (for a first-time poster:smile:): is there something in the link ehild gave that you don't understand ? Do you know what you need from there ?
     
  8. May 24, 2017 #7
    eaaaa.jpg
    here is my problem but in the other form is resolved yet
     
  9. May 24, 2017 #8

    BvU

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    And what does the theory say about z on ##R_{xy}## ?
     
  10. May 24, 2017 #9
    Rxy is the projection of S in the plane xy.
     
  11. May 24, 2017 #10

    BvU

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    So ##
    R_{xy}## is a quarter circle (not just the circumference).
    But ##z=0## in the plane xy, so what do you do with that ##R\over z## ? Where does this theoretical formula come from ? Is it applicable ?

    Would you be interested to follow the path ehild and I learned long ago and proposed here in #2 and #3 ?
     
  12. May 24, 2017 #11
    yes i am interested. here is the resolution by put the sphere in parametric form r(u,v)
    exercise.jpg
     
  13. May 24, 2017 #12
    my problem is to resolve it with surface in this form: G(x,y,z)=x^2+y^2+z^2-R^2=0
    the theory is
    upload_2017-5-24_14-59-5.png
    or equivalent
    upload_2017-5-24_14-59-35.png
    or
    upload_2017-5-24_14-59-48.png
    the real problem is in the f(x,y,z(x,y)) or the other equivalent
     
  14. May 24, 2017 #13

    BvU

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    Excellent work.

    So -- if the theory is correct and applicable -- we (or rather, your helpers) are back to understanding what is needed for the alternative route.
    What is meant with ##\left | dG\over dz\right |## in the formula ?
    Does it perhaps mean you need to evaluate this factor in the numerator at ##z= \sqrt{R^2 - (x^2+y^2)\,}\ ## ?

    [edit] oopsed & fixed.
     
  15. May 24, 2017 #14
    I will try it, thanks!
     
  16. May 24, 2017 #15
    but ##x²+y²=R²## so ##z= \sqrt{R^2 - (x^2+y^2)\,}=0##
     
  17. May 24, 2017 #16
    this way of solve it is so difficult
     
  18. May 24, 2017 #17

    BvU

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    No. That is on the rim of the circle, not in the interior. The projection of S on the xy plane is the whole quarter circle, not just the edge.
     
  19. May 24, 2017 #18
    could you write it ?
     
  20. May 24, 2017 #19

    BvU

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    So your bounds are e.g. 0-1 for x, 0-##\sqrt{1-x^2\,}## for y
     
  21. May 24, 2017 #20
    o-R for x right?
     
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