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Double integral of arctan in polar coordinates

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral using polar coordinates:

    ∫∫arctan(y/x) dA

    Where R={ (x,y) | 1≤ x2 + y2 ≤ 4, 0≤y≤x

    2. Relevant equations

    X=rcos(T)
    Y=rsin(T)
    r2=x2 +y2

    3. The attempt at a solution

    First thing was drawing a picture of R, which I think looks like a ring 1 unit thick centered at the origin and with a radius between 1 and 2. Since y≤x, the ring is sliced through the middle by the line y=x, and only the lower half in quadrants 1, 3, and 4 is a part of R.

    I rewrote the region as R= {(r, T) | -3pi/4 ≤ T ≤ pi/4, 1 ≤ r ≤ 2}.

    After evaluating r*arctan(tan(T))drdT at the above limits, I end up with 3, while the book says the answet is 3pi2/64. I think I am translating the region incorrectly, but don't see how else to describe it.

    Thanks for the help!
     
    Last edited: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    SammyS

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    Hello LunaFly. Welcome to PF !

    Notice that 0 ≤ y ≤ x , so that not only is x ≥ y, but also y ≥ 0 and x ≥ 0.

     
  4. Mar 14, 2013 #3
    Solved!

    Ah overlooking that small detail really messed up my work!

    Thanks for pointing out my mistake :approve: !
     
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