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Double integral over a region needing polar coordinates.

  1. Nov 6, 2013 #1
    1. Evaluate the double integral ∫∫arctan(y/x) dA by converting to polar coordinates over the Region R= { (x,y) | 1≤x^2+y^2≤4 , 0≤y≤x }


    My attempt at solving

    Converting to polar using x=rcosθ and y=rsinθ I get
    ∫∫arctan(tan(θ))r drdθ


    I understand that I have to integrate first with respect to r and then with respect to θ but I'm not sure what the region is supposed to look like over which I am integrating so I don't know what my bounds should be. I understand that 1≤x^2+y^2≤4 is basically 2 circles, one with radius 1, and one with radius 2 and my region is between them. I don't know how 0≤y≤x affects it though, I thought it would just make it the top right portion of the circle but thats incorrect.
    Any help is greatly appreciated, thanks.
     
    Last edited: Nov 6, 2013
  2. jcsd
  3. Nov 6, 2013 #2

    tiny-tim

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    Hi Tropicalism! Welcome to PF! :smile:
    "0≤y and 0≤x" would be top right portion (the first quadrant :wink:)

    "0≤y≤x" is the half of it under the diagonal x=y :smile:
     
  4. Nov 6, 2013 #3
    Oh, ok thank you. So using polar that portion would end at θ=∏/4 correct?
     
  5. Nov 6, 2013 #4

    tiny-tim

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  6. Nov 6, 2013 #5
    Thanks! Thats all I needed.
     
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