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Math Help - need help with double integral

  1. #1
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    need help with double integral

    hey guy, I'm having a problem with double integral integration. Can you guys tell me where I went wrong?




    The equation is \int_0^1\int_0^y\frac{y^3}{x^2+y^2}dxdy

    let U = x+y
    \frac{dU}{dx} = 1
    \int_0^1\int_0^y\frac{y^3}{U^2}dUdy
    \int_0^1[\frac{-y^3}{x+y}]_0^y dy
    \int_0^1 \frac{y^2}{2} dy
    [\frac{y^3}{6}]_0^1 = \frac{1}{6}

    However, I seem like this answer is wrong, and the correct answer should be \frac{pi}{12}
    Can anyone point out what I did wrong here?

    Best Regards
    Junks
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  2. #2
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    Re: need help with double integral

    $(x+y)^2 = x^2 + y^2$

    Really? Try it again.
    Thanks from junkwisch
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  3. #3
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    Re: need help with double integral

    ahh, I see, I'll let U=x^2 + y^2
    .................................................. ..........................................

    So I let U=x^2+y^2
    \int_0^1\int_0^y\frac{y^3}{2*x*U}
    \int_0^1[\frac{y^3}{2*x}*log(x^2+y^2)]_0^y
    \int_0^1\frac{y^2}{2}*log(2y^2)
    let U=log(2*y^2) and \frac{dV}{dy}=\frac{y^2}{2}
     \frac{dU}{dy}=\frac{2}{y} and V=\frac{y^3}{6}
    using this rule UV-\int \frac{dU}{dy}*\frac{dV}{dy}dy
    [\frac{y^3}{6}*log(2*y^2)-\int y dy]_0^1
    \frac{1}{6}*log(2)-\frac{1}{2} the answer turn out to be -0.38, but the correct answer seem to be pi/12 or 0.26. What did I do wrong in here?
    Last edited by junkwisch; March 29th 2014 at 07:11 PM.
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  4. #4
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    Re: need help with double integral

    \int_0^1\int_0^y\frac{y^3}{x^2+y^2}dxdy=

    \int_0^1 y^3 \int_0^y \dfrac{1}{x^2+y^2}~dx~dy=

    \int_0^1 y^3 \left(\dfrac{\arctan(\frac{x}{y})}{y}|_0^y\right)~  dy=

    \int_0^1 y^2 \left(\arctan(1)-\arctan(0)\right)~dy=

    \int_0^1 y^2 \dfrac{\pi}{4}~dy=

    \dfrac{\pi}{4}\left( \dfrac{y^3}{3}~|_0^1\right)=

    \dfrac{\pi}{12}
    Thanks from junkwisch
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