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Math Help - [SOLVED] Another improper integral

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Another improper integral

    \int \int_D \frac{dxdy}{\sqrt{xy}} where D = \left\{x+y\leq1, x> 0, y> 0\right\}

    \int_b^1 \left(\int_b^{1-x} \frac{dy}{\sqrt{xy}}\right)dx = 2\int_b^1 \frac{1}{x}(\sqrt{x-x^2} -\sqrt{bx})dx

    I can't solve the last integral, and the answer I get from Mathematica does not work for these limits. So I suspect the above isn't correct.

    EDIT: Here's another tricky improper integral:

    \int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}} where \sqrt{x^2+y^2} < z < 1

    EDIT: And another one:

    \int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}
    Last edited by Spec; August 13th 2008 at 03:46 PM.
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  2. #2
    Eater of Worlds
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    Are you sure that isn't dydx instead of dxdy?. The way it is you have a dependent limit.
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  3. #3
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    Quote Originally Posted by Spec View Post
    I can't solve the last integral, and the answer I get from Mathematica does not work for these limits.
    Hey Spec. This is what I get with Mathematica 6: What's wrong with it?

    \text{If}\left[\left(\text{Re}\left[\frac{b}{1-b}\right]\geq 0\&\&\frac{b}{-1+b}\neq 0\right)\left\|\text{Re}\left[\frac{b}{1-b}\right]\leq -1\right\|\text{Im}\left[\frac{b}{1-b}\right]\neq 0]\right]

     <br />
\frac{-\frac{4}{\sqrt{\frac{1}{b}}}+\left(2 \left(\sqrt{-1+b} \sqrt{b} \left(-1+2 \sqrt{-\frac{b}{-1+b}}\right)+\text{Log}\left[\sqrt{-1+b}+<br />
\sqrt{b}\right]\right)\right)}{\left(\sqrt{-1+b} \sqrt{\frac{1}{b}} \sqrt{\frac{b}{1-b}}\right)}

    Also, if you want to cut-and-paste Mathematica as latex code, select the code in Mathmatica, then choose Edit/Copy as/Latex (with some minor limitations). For example, I didn't type in any of that code above although I did have to re-format it slightly because of some limitations on latex length in here.
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  4. #4
    Senior Member Spec's Avatar
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    The integral (and answer) looks way too complicated, and we're supposed to be able to solve all these problems without using any tools.
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  5. #5
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    Quote Originally Posted by Spec View Post
    \int \int_D \frac{dxdy}{\sqrt{xy}} where D = \left\{x+y\leq1, x> 0, y> 0\right\}

    =\int_0^1 \frac{2}{\sqrt{x}} [ \sqrt{y}]_0^{1-x} \ dx=2\int_0^1 \sqrt{\frac{1-x}{x}} \ dx=4\int_0^{\frac{\pi}{2}} \cos^2 \theta \ d \theta = \pi. \ \ \ \square

    in the last integral we put x=\sin^2 \theta.


    \int \int \int_D \frac{dxdydz}{\sqrt{x^2+y^2+z^2}} where \sqrt{x^2+y^2} < z < 1

    =\int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_0^{\sec \phi} \rho \sin \phi \ d \rho \ d \phi \ d \theta=\pi \int_0^{\frac{\pi}{4}} \sec^2 \phi \ \sin \phi \ d \phi = \pi [\sec \phi]_0^{\frac{\pi}{4}}=\pi(\sqrt{2} \ - \ 1). \ \square

    note that in spherical coordinates z=\sqrt{x^2+y^2} becomes \phi = \frac{\pi}{4} and z=1 becomes \rho=\sec \phi.


    \int \int_{R^2} \frac{xdxdy}{(x^2+y^2)^2 +1}

    = \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square

    note that \int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr is convergent.
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  6. #6
    Senior Member Spec's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    = \int_0^{\infty} \int_0^{2 \pi} \frac{r^2 \cos \theta}{r^4 + 1} \ d \theta \ dr = \int_0^{\infty} \frac{r^2}{r^4 + 1} [\sin \theta]_0^{2 \pi} \ dr=0. \ \ \ \square

    note that \int_0^{\infty} \frac{r^2}{r^4 + 1} \ dr is convergent.
    Does that matter though? Since your multiplying it with zero either way. I actually came that far myself, but struggled with the calculation of \int_0^{\infty} \frac{r^2dr}{r^4 + 1} (I tried to integrate it before d\theta, and thus didn't notice that the d\theta integral was zero).
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