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Math Help - Double Integrals - Order of Integration

  1. #1
    Member billym's Avatar
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    Double Integrals - Order of Integration

    Find the area of A by evaluating the double integral

    \int\int_A \, dxdy

    where the region A is bounded by the curve y=x^2-x and the line y=2

    Is this correct:

    \int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx

    Since it says evaluate \int\int_A \, dxdy, do you reckon I have to change the order of integration?

    If so, I have:

    \int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy

    Is this ridiculous?
    Last edited by billym; November 25th 2009 at 09:59 AM.
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  2. #2
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    Quote Originally Posted by billym View Post
    Find the area of A by evaluating the double integral

    [tex] \int\int_A \, dxdy

    where the region A is bounded by the curve y=x^2-x and the line y=2

    Is this correct:

    \int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx

    Since it says evaluate \int\int_A \, dxdy, do you reckon I have to change the order of integration?

    If so, I have:

    \int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy

    Is this ridiculous?

    It looks pretty absurd: better, just change dxdy by dydx and do the first, much easier, integration.

    Tonio
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