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Factorising Integration

  1. Apr 12, 2007 #1
    [​IMG]

    [​IMG]
     
  2. jcsd
  3. Apr 12, 2007 #2

    Integral

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    Just do the substitution.

    in part b, let [itex] k^3 = \rho ^2 [/itex]

    Solve the resulting quadratic.
     
  4. Apr 12, 2007 #3
    [​IMG]

    ??? :S
     
  5. Apr 12, 2007 #4
    k^3 = p^2 implies k^(3/2) = p.
     
  6. Apr 12, 2007 #5
    I see now, so silly. Thanks
     
  7. Apr 13, 2007 #6
    Still cannot get answer.

    [​IMG]
     
  8. Apr 13, 2007 #7

    dextercioby

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    You've gotten "p", now you have to find "k".
     
  9. Apr 13, 2007 #8
    How is that going to lead me to the answer? :redface:
    If I substitute "p" back to "k". I don't see how I can get 0.44
     
  10. Apr 13, 2007 #9
    What is an expression for k in terms of p?

    And you'll get an extraneous solution, but which one is right will be fairly clear when you get them. Look at bounds of your integration. Then look at the k's you get.
     
  11. Apr 13, 2007 #10
  12. Apr 13, 2007 #11

    HallsofIvy

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    Do you see an error here?
     
  13. Apr 13, 2007 #12
    [​IMG]

    I hope...


    I don't quite understand, how i'm I to know 0.29 instead of 1.71 was to be used?
     
    Last edited: Apr 13, 2007
  14. Apr 13, 2007 #13
    What is 1.79^(2/3)? Why is it that k clearly can't take on that value?
     
  15. Apr 13, 2007 #14
    Because of: [​IMG]?
     
  16. Apr 13, 2007 #15
    You've pretty much got the right idea. When you split the integral from 0 to 1 into an integral from 0 to k and an integral from k to 1, you tacitly assumed that k was between 0 and 1. Otherwise, what you did wouldn't be valid.
     
  17. Apr 13, 2007 #16
    Ok, thanks Mystic998 and everyone else who helped out.
     
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