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Math Help - Iterated Integrals

  1. #1
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    Iterated Integrals

    Hey Everyone,

    Suppose I have this function f(x,y) = 8xy

    I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

    I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

    Thanks again!
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  2. #2
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    Quote Originally Posted by JoAdams5000 View Post
    Hey Everyone,

    Suppose I have this function f(x,y) = 8xy

    I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

    I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

    Thanks again!
    you need to rotate your curve to get rid off the term xy and get a standard form of an ellipse. the substitution x=u+v, \ y=u-v will do that for you.

    you can also directly transform your curve to a circle instead of a (standard) ellipse and then a circle. the substitution x=u+\sqrt{3}v, \ y=u-\sqrt{3}v will do the job.
    Last edited by Krizalid; July 6th 2009 at 04:19 PM.
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  3. #3
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    Quote Originally Posted by JoAdams5000 View Post
    Hey Everyone,

    Suppose I have this function f(x,y) = 8xy

    I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4

    I am having a little trouble constructing the integrands. I believe that -1 < x < 1, but I am not positive about what the range of y is.

    Thanks again!
    How about just solving it as is where is first as in 2\int_0^1 \int_{g_1(y)}^{g_2(y)} 8 xy\,dxdy where g(y)=\frac{-y\pm \sqrt{3-3y^2}}{2}? Pretty sure that's correct but haven't worked through it yet. If it's not, try and fix it please.
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  4. #4
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    Quote Originally Posted by JoAdams5000 View Post
    Hey Everyone,

    Suppose I have this function f(x,y) = 8xy

    I want to evaluate the double integral of this function over the region enclosed by the curve x^2 + y^2 + xy = 3/4
    Put x=\frac{u-\sqrt{3}v}{\sqrt{6}},\,y=\frac{u+\sqrt{3}v}{\sqrt{  6}} and your region will be transfered to u^{2}+v^{2}\le \frac{3}{2} and then apply the usual polar transformation.
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  5. #5
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    Thanks everyone.

    I tried doing it all three ways and I either cannot finish the integration or I get a negative answer, which I do not think is correct.

    If anyone does calculate this and gets a plausible answer, please let me know!

    Thanks again
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  6. #6
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    UPDATE!

    I tried using a change of variables and then solving along with jacobian value and got an answer of 8/sqrt(3). Whether this is right or wrong, it is anybody's guess.

    I am almost positive that it should be possible by find the range of x and y and solving the integral, without making any substitutions.

    I found it to be -1<x<1 and -x/2 - sqrt(3-3x^2)/2 < y < -x/2 + sqrt(3-3x^2)/2.

    However, when I try to solve this...i get -sqrt(3)pi/2...

    Any ideas? or do you see any errors i may have made?
    Thanks
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  7. #7
    MHF Contributor matheagle's Avatar
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    Since 8xy is negative and positive (in your region) you cannot claim that
    \int\int 8xy dxdy>0, so a negative answer is not impossible.
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  8. #8
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    Quote Originally Posted by matheagle View Post
    Since 8xy is negative and positive (in your region) you cannot claim that
    \int\int 8xy dxdy>0, so a negative answer is not impossible.
    Thanks! I had a feeling that did not make sense. However, I don't think I made any integration mistakes so that means the region i am integrating on is incorrect...but then i am pretty sure that is right also. I guess it is time to do an extensive check of all my calculations
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