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Double integrals using polar co-ordinates

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫∫e-(x^2+y^2) dA
    R

    Where R is the region enclosed by the circle x2+y2=1

    First thing I did was graph the region where the function was enclosed. I saw that they didnt give a limitation to where the circle lied. So I automatically knew that d(theta) would lie on the interval 0≤theta≤2π.

    Here is the part where things get a bit tricky for me, I need to find what dr is

    So my thought process automatically told me to change x2+y2=1

    Into polar coordinates.

    Using the following relations x=rcos(theta); y=rsin(theta) and plugging in I get the relation

    r^2=1

    So I would think dr would lie on the interval from -1≤r≤1

    However what the book says is that r lies on 0≤r≤1.

    This I do not understand.

    Thanks

    Higgenz

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 28, 2012 #2

    Dick

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    r generally chosen to be a nonnegative number. Then 0<=r<=1 and 0<=theta<2*pi covers the unit disk once. If you let -1<=r<=1 then you have covered the unit disk twice. Changing theta->theta+pi and r->(-r) wouldn't change x and y.
     
  4. Oct 28, 2012 #3
    Wow thanks dick. However, what if for example we had a the same circle

    X^2+y^2=4

    Would then go from 0≤r≤2 ?
     
  5. Oct 28, 2012 #4

    Dick

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    Sure. Same thing.
     
  6. Oct 28, 2012 #5
    Awesome thank you
     
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