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Math Help - Changing the order of integration

  1. #1
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    Changing the order of integration

    Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)

    Evaluate
     \int_{x=0}^2\int_{y=\frac{x}{2}}^1 2xy^2 dy dx

    first directly then changing the order of integration
    I did the first part and got 0.8 as the answer

    just for the second I'm having trouble

    I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

    \int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy

    but the value of that doesnt agree, can anyone explain how to do the limits correctly?

    cheers

    Simon
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  2. #2
    Senior Member Peritus's Avatar
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    <br /> <br />
\int\limits_0^1 {\int\limits_0^{x = 2y} {2xy^2 dxdy} } <br /> <br />
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thelostchild View Post
    Hey I'm having some trouble with multiple integrals if anyone would care to explain (I think I'm just not using the correct region!)



    I did the first part and got 0.8 as the answer

    just for the second I'm having trouble

    I think the surface being integrated over the the triangle bounded by the lines y=x/2 x=0 and y=1 but I may be wrong which gives

    \int_{y=0}^1\int_{x=2y}^2 2xy^2 dx dy

    but the value of that doesnt agree, can anyone explain how to do the limits correctly?

    cheers

    Simon
    0\leqslant{x}\leqslant{2} and \frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqs  lant{2y}\leqslant{2}}. Stringing these two inequalites together gives

    0\leqslant{x}\leqslant{2y}\leqslant{2}

    From there Peritus's answer should be more apparent.
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by Mathstud28 View Post

    0\leqslant{x}\leqslant{2} and \frac{x}{2}\leqslant{y}\leqslant{1}\implies{x\leqs  lant{2y}\leqslant{2}}. Stringing these two inequalites together gives

    0\leqslant{x}\leqslant{2y}\leqslant{2}

    From there Peritus's answer should be more apparent.
    But be careful 'cause, this method not always works.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    But be careful 'cause, this method not always works.
    Really? I have never encountered a case where it hasn't. Would you mind giving me an example of one please?
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  6. #6
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    Try it with \int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Try it with \int_0^1\int_{x^3}^{\sqrt[3]x}dy\,dx.
    Thank you, I will report back later when I have had time to look at it.
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  8. #8
    Moo
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    Also be careful that reversing integration order is submitted to the condition that \int \int \left|f(x,y)\right| ~ dx ~ dy is a finite value.
    See Fubini's theorem - Wikipedia, the free encyclopedia for further information (advanced calculus imo)
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