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How to integral ? How can they ignore that variable?

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data
    I don't understand case 1. In case 2 that is normal integration , isn't?
    Then why case 1 can simply ignore the r ? I don't think r is constant.


    2. Relevant equations



    3. The attempt at a solution
    Please guide. Thanks
     
    Last edited by a moderator: Jul 25, 2013
  2. jcsd
  3. Jul 25, 2013 #2

    Mark44

    Staff: Mentor

    Please upload your attachment again, but oriented so people can read it.
     
    Last edited: Jul 25, 2013
  4. Jul 25, 2013 #3
    This way? No sure is that correct .
     

    Attached Files:

  5. Jul 25, 2013 #4

    Mark44

    Staff: Mentor

    It's upside-down now. You can edit your post and try uploading the attachment again.
     
  6. Jul 25, 2013 #5
    Sorry, I don't really know how to use it.
     

    Attached Files:

  7. Jul 25, 2013 #6

    LCKurtz

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    Are you talking about the equation$$
    \frac m r \frac{d(r^2\dot \theta)}{dt}=0$$giving ##r^2\dot \theta=l##? If so, they didn't ignore the ##\frac m r##. They multiplied both sides by its reciprocal before integrating.
     
  8. Jul 25, 2013 #7
    I'm not quite sure what you mean. Nobody said r was constant. But the fact that [itex]\frac{d}{dt}(r^2\theta') = 0 [/itex] does imply that [itex]r^2\theta' = C [/itex] is constant (why?).
     
  9. Jul 25, 2013 #8
    (1/r)(dr/dt)= 0 , (dr/dt)= 0
    ∫ dr = ∫ 0 dt
    The answer should be r = constant.
    Isn't ?
    Can the answer be ∫(1/r)dr= ∫dt .
    Then ln(r/ro)= t. Integral from ro to r , and integral from 0 to t. ro is original value r.
     
  10. Jul 25, 2013 #9

    LCKurtz

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    I thought you said you were puzzled about 1, not 2.


    Yes, that's right.

    ##\int \frac 1 r\, dr = \int 0\, dt##, not ##\int 1\, dt##.
     
  11. Jul 26, 2013 #10
    I got it. Thanks much
     
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