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Math Help - Evaluate the following double integrals

  1. #1
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    Evaluate the following double integrals

    I swear, our Prof. must get shot at at least once a year for some of the crazy assignments he gives out. At least one student per year in his class must go insane. I digress, however.

    These two double integrals (the material for which I'm trying to catch up to) are a lot more difficult than they look (or I made a mistake somewhere). I have the final answers due to an online website, but it doesn't give processes for definite integrals (which I'm trying to work out right now).

    a) \int_0^4 \int_{\sqrt{y}}^2 \frac{1}{1+x^3} dxdy
    Ans. \approx 0.732408

    b) \int_0^1 \int_x^1 \frac{x}{1+y^6} dydx
    Ans. \approx 0.1309

    I'll put up work on this a bit later when I have more, but any help that could be provided would be appreciated.
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  2. #2
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    Quote Originally Posted by Runty View Post
    I swear, our Prof. must get shot at at least once a year for some of the crazy assignments he gives out. At least one student per year in his class must go insane. I digress, however.

    These two double integrals (the material for which I'm trying to catch up to) are a lot more difficult than they look (or I made a mistake somewhere). I have the final answers due to an online website, but it doesn't give processes for definite integrals (which I'm trying to work out right now).

    a) \int_0^4 \int_{\sqrt{y}}^2 \frac{1}{1+x^3} dxdy
    Ans. \approx 0.732408

    b) \int_0^1 \int_x^1 \frac{x}{1+y^6} dydx
    Ans. \approx 0.1309

    I'll put up work on this a bit later when I have more, but any help that could be provided would be appreciated.
    The trick with these is to reverse the order of integration. Start by drawing a diagram of the region over which the integral occurs. For (a), the region described by \sqrt y\leqslant x\leqslant 2,\ 0\leqslant y\leqslant 4 is the region under the parabola y=x^2 between x=0 and x=2. So it could also be described as the region 0\leqslant y \leqslant x^2,\ 0\leqslant x\leqslant 2.

    That means that when you switch the order of integration, the integral becomes \displaystyle\int_0^2\!\!\int_0^{x^2}\!\!\!\frac1{  1+x^3}\,dydx, which you should find much easier to do. (I make it \frac13\ln9.)

    I haven't looked at (b), but I think that it will almost certainly respond to the same technique.
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  3. #3
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    Wow, and I was doing that one the hard way. That just shows how out of this sort of thing I am. I definitely picked the wrong degree; I'm better at number crunching and data entry, not critical thinking.

    EDIT: Would the region for b) be as follows for reversing the order of integration?
    0\leq x\leq y,\ 0\leq y\leq 1
    Last edited by Runty; November 29th 2010 at 01:30 PM.
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  4. #4
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    Quote Originally Posted by Runty View Post
    EDIT: Would the region for b) be as follows for reversing the order of integration?
    0\leq x\leq y,\ 0\leq y\leq 1
    Correct. In problems such as these, I always find it helpful to draw the region enclosed.
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