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What is the mistake here

  1. Jul 1, 2011 #1
    why does the prof wrote a mistake there and wrote 2cos theta in there
    ??

    (i didnt typed the question correctly,the original is
    i need to calculate the integral on the volume enclosed by z>=0 and the sphere wich is written in the photo.)
    24oq2ww.jpg
    how mathematicky can i get this transition

    how to get this expression
    from cartesian to polar and get the 2 cos teta on the interval

    [TEX]\rho\le 2\cos \phi[/TEX]

    ??
     
  2. jcsd
  3. Jul 1, 2011 #2

    tiny-tim

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    hi nhrock3! :smile:

    if the origin is O, and if the centre of the sphere is C, and if P is a point on the sphere,

    what are the angles of triangle OCP, and what is the length of OP ? :wink:
     
  4. Jul 1, 2011 #3
    OP^2=0C^+PC^2
    how it changes the intervals?
     
  5. Jul 1, 2011 #4

    tiny-tim

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    no, it's not a right-angle
     
  6. Jul 1, 2011 #5

    SammyS

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    What are the lengths of OP & PC ?

    Isn't φ the angle from the z axis to OP ?

    Use the law of cosines to answer tiny-tim's question.
     
  7. Jul 2, 2011 #6

    i cant imagine the angle


    Isn't φ the angle from the z axis to OP ?


    could you draw it please
     
  8. Jul 2, 2011 #7

    tiny-tim

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    C is the centre, so CO = CP = r = 1.

    Angle COP = φ.

    So how long is OP?​
     
  9. Jul 2, 2011 #8

    micromass

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    Or just use that [itex]x^2+y^2+z^2=\rho^2[/itex] and [itex]z=\rho \cos\phi[/itex] to transform

    [tex]x^2+y^2+z^2=2z[/tex]

    into an equation that uses only rho and phi.
     
  10. Jul 2, 2011 #9
    OP^2=CO^2+CP^2-2OC*PCcos(180-2φ)

    what to do now?
     
  11. Jul 2, 2011 #10

    tiny-tim

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    well, now put the numbers in :confused:
     
  12. Jul 2, 2011 #11
    ok i got it
    if we try and solve it in a different way.
    but if we do variable change
    x=u y=v z-1=w
    then the jacobian is 1
    and the integral is a ball of radius from 0 to 1

    unlike here where our radius is from 0 to 2cos

    why??
     
  13. Jul 2, 2011 #12

    SammyS

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    Placing the origin at the center of the sphere naturally makes the limits of integration simple for a sphere. I hope you changed the integrand accordingly.
    z → w+1
     
  14. Jul 2, 2011 #13

    tiny-tim

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    no, nhrock3, you haven't got it …

    you still haven't a clue why your prof did that …
    your prof is a clever guy who knows how best to teach this subject (and who knows what's coming up in the exams :rolleyes:)

    it is not clever for you to give up on his method and to "try and solve it in a different way" :redface:

    (and presumably you still think your prof wrote a mistake?)

    start again …​
    in other words, put CO = CP= PC = 1 …

    what do you get?
     
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