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Math Help - double integral

  1. #1
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    double integral

    I = ∫∫R (x + y)dA where R is the region bounded by the curve y = x and the line y = 1.
    (a) Calculate I by integrating first in y and then in x.
    (b) Calculate I by integrating first in x and then in y.

    This is what I got for (a): Limits for y are x =< y =< 1, making this a parabola from origin to y=1. solving x=1 (upper limit of y) gives the limits for x -1=<x=<1.

     \int_{-1}^{1}\int_{x^2}^{1} (x+y)

     \int_{-1}^{1} (xy+ \frac{y^2}{2})\bigg|_{x^2}^{1}


     \frac{1}{2} \int_{-1}^{1} (-x^4 - 2x^3 + 2x +1)

     \frac{1}{2} (\frac{-x^5}{5} - \frac{x^4}{2} + x^2 +x)\bigg|_{-1}^{1}

     =\frac{4}{5}

    which I hope is correct!

    I'm stuck on (b)
    What should I make the limits of x & y to?
    are the limits for x: -1 \leq x \leq \sqrt{y}? This is from solving x = y, making x=\sqrt{y}
    But what do I make the limits for y? If I use 0 \leq y \leq 1 I get a different answer from (a). I get 13/20.

    where am I going wrong?
    Last edited by Dr Zoidburg; May 4th 2010 at 06:31 PM.
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  2. #2
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    x^2=y ~\implies~ |x| = \sqrt{y} ~\implies~ x = \pm \sqrt{y}

    So, this means that x=\sqrt{y} where x is positive x=-\sqrt{y} where x is negative.

    Therefore, the endpoints will be

    \int_0^1 \int_{-\sqrt{y}}^{\sqrt{y}} (x+y) \, dx \, dy
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  3. #3
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    D'oh! It's so obvious. Thanks heaps for that. It was bugging me no end I've got this assignment due in a couple of days.
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