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Evaluating a double integral

  1. Jan 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral (x^2+y^2)arctan(y/x) for 0<y<a and 0<x<(a^2-y^2)^0.5.

    2. Relevant equations



    3. The attempt at a solution

    I tried changing the order of integration to get the integral (x^2+y^2)arctan(y/x) for 0<x<a and 0<y<(a^2-x^2)^0.5. I noticed that this was a quarter of a circle.

    I tried then taking x^2 out of the dy integral and into the dx one. The dy integral is then (1+(y/x)^2)arctan(y/x). I'm not sure what to do now.
     
    Last edited: Jan 23, 2012
  2. jcsd
  3. Jan 22, 2012 #2
    Do you use the fact that x^2+y^2=a^2? I can then take the constant a^2 in front of the integral, so I only need to integrate arctan(y/x).dy.
     
  4. Jan 23, 2012 #3

    Mark44

    Staff: Mentor

    Is the last inequality supposed to be 0 < x < √(a2 - y2)?


    Rather than just changing the order of integration, I think the best plan is change to a polar integral.
     
  5. Jan 23, 2012 #4
    Yes, I've changed that now- thanks for pointing out the error.

    So is it a^2∫θ.dθ∫r.dr?

    How would I change the limits?
     
  6. Jan 23, 2012 #5
    So would the limits be 0<r<a and -π/2<θ<π/2?
     
  7. Jan 23, 2012 #6
    Would someone mind checking whether these limits are correct? :-)

    Thanks in advance.
     
  8. Jan 23, 2012 #7

    I like Serena

    User Avatar
    Homework Helper

    Hi Lucy Yeats! :smile:

    It should be x^2+y^2=r^2.
    It is not equal to a^2, since you integrate (x,y) over the surface of the circular disk and not just the boundary.

    Your angle θ should run from 0 to pi/2, since you only integrate the first quadrant.
    For θ<0 you would get negative y, but your problem statement says y>0.
     
  9. Jan 27, 2012 #8
    I've got it now, thanks everyone! :-)
     
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