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Double Integral Cartesian to Polar Coordinates

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Use polar coordinates to evaluate:
    sqrt(2)0sqrt(4-y2)y 1/(1+x2+y2) dxdy


    2. Relevant equations



    3. The attempt at a solution
    I graphed it and I see r is the part of the elipse sqrt(4-y2) and goes from 0 to ∏/4. I'm not sure how to make the bounds for r or how to covert the function to polar. Does it just convert to 1/(1+r2), then add the conversion factor r?
    ∏/40?? 1/(1+r2) r drdθ
     
  2. jcsd
  3. May 6, 2012 #2

    sharks

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    Gold Member

    Use LaTeX formatting to make your integrals clearer and easier to understand.

    Your double integral looks like this:
    [tex]\int^{y=\sqrt 2}_{y=0} \int^{x=\sqrt {4-y^2}}_{x=y} \frac{1}{1+x^2+y^2} \,.dxdy[/tex]
    You have to plot the limits to see the area required and hence deduce the correct polar limits.
    First, convert: [itex]x=\sqrt {4-y^2}[/itex] into a simpler form. This gives you: [itex]x^2+y^2=4[/itex], which is a circle with center (0,0) and radius = 2. The other limits are obvious.

    I have attached the graph. If you are having trouble deducing the region required from the limits, save the graph, edit/shade it to show the required area and then attach it to your reply.
     

    Attached Files:

    Last edited: May 6, 2012
  4. May 6, 2012 #3
    I'm just not sure what to put for the r limits. I know it is 0 to the circle x2+y2=4. So is it just 0 to 2?
     

    Attached Files:

  5. May 6, 2012 #4

    sharks

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    Gold Member

    You're correct. :smile:
     
  6. May 6, 2012 #5
    Thanks!!
    I think I figured it all out now. Thanks again.
     
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