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Math Help - Double integral

  1. #1
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    Double integral

    \displaystyle A=\int_{-1}^0\int_0^{(x+1)^2}f(x,y) \ dy \ dx + \int_0^1\int_0^{(x-1)^2} f(x,y) \ dy \ dx<br />

    By reversing the order of integration, express A as a single iterated integral.

    Would it be like this

    \displaystyle A=\int_{0}^1\int_{\sqrt{y}-1}^{0}f(x,y) \ dx \ dy + \int_0^1\int_0^{\sqrt{y}+1} f(x,y) \ dx \ dy
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  2. #2
    A Plied Mathematician
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    On the second integral, I think you need to take the negative square root. That is, I think you should have

    \displaystyle\int_{0}^{1}\int_{0}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    On the second integral, I think you need to take the negative square root. That is, I think you should have

    \displaystyle\int_{0}^{1}\int_{0}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.
    The answer is apparently

    \displaystyle\int_{0}^{1}\int_{1-\sqrt{y}}^{\sqrt{y}-1}f(x,y)\,dx\,dy

    I don't understand why.
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  4. #4
    A Plied Mathematician
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    If the book had said

    \displaystyle\int_{0}^{1}\int_{\sqrt{y}-1}^{-\sqrt{y}+1}f(x,y)\,dx\,dy,

    I would have believed it.
    Last edited by Ackbeet; November 6th 2010 at 07:27 PM. Reason: Comma instead of period.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    If the book had said

    \displaystyle\int_{0}^{1}\int_{\sqrt{y}-1}^{-\sqrt{y}+1}f(x,y)\,dx\,dy.

    I would have believed it.
    Yeah Ackbeet, I managed to now get the same answer as you. Maybe the answer at the back of the book is wrong. Thanks anyways
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  6. #6
    A Plied Mathematician
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    I think the book is wrong. You're welcome!
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