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Change to polar coordinates integration Problem

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Integrate y/(x^2+y^2) for x^2+y^2<1 and y> 1/2 ; use change of variables to polar coordinates

    2. Relevant equations

    THe above



    3. The attempt at a solution

    the variables transform as

    y=rsinz
    x=rcosz, where z is an angle between pi/6 and 5*pi/6 = which is the ArcSin of 1/2

    That is sinz > 1/2 means pi/6 < z < 5*pi/6,

    The Integral in polar coordinates becomes

    Integral of [ r*sinz / { (r*sinz)^2 + (r*cosz)^2 } ] * r drdz

    This simplifies to

    Integral of [r*sinz / { r^2 * 1 } ]*r drdz, or

    =Integral [ sinz drdz ]

    Now y > 1/2 means that r must be between [ (1/2)^ + (cosz)^2 ] ^ (1/2) and 1

    by the Pythagorean Theorum.

    So i end up with the double integral

    Integral rsinz dz evaluated at r = 1, minus same evaluated at
    r = [ (1/2)^ + (cosz)^2 ] ^ (1/2)

    So I then get Integral [ 1 - [ (1/2)^ + (cosz)^2 ] ^ (1/2) ] * sinz dz.

    The book gives the answer (sqrt 3 ) - pi/3, that is after evaluating what I would think would be the above from z = pi/6 to 5*pi/6.

    The integral I came up with after substituting the range of r stated as functions of z, does not seem to integrate and must be wrong -

    Can anyone tell me where I went off track ?

    Would be much appreciated - hours on this, and cannot get it.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 4, 2012 #2

    SammyS

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    .

    Hello Delphi27. Welcome to PF !

    Why not express this in terms of sin(z), rather than cos(z) ?

    You know that y = r sin(z), right?

    So y = 1/2 becomes r sin(z) = 1/2.

     
  4. Aug 4, 2012 #3
    [itex] y > \frac {1} {2} [/itex] means [itex]r sin z > \frac {1} {2}[/itex]
     
  5. Aug 4, 2012 #4
    Volko and SammyS --- thanks so much - you guys got me on the right track !!
     
  6. Aug 4, 2012 #5

    SammyS

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    You're very welcome !
     
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