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Math Help - The measure of a subset of the real plane given by inequalities

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    The measure of a subset of the real plane given by inequalities

    I have to calculate the Lebesgue measure of

    \left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}

    for every k\in\mathbb{Z}
    After the substitution I get \int_U r drd\alpha dz, where

    U=\left\{0<r<1,\,\alpha\in(0,2\pi),\,(1-r^2)^kz^2<r^4\cos2\alpha\right\}.

    I can't calculate it. Everything I try leads to difficult integrals. Could you please give me an idea of what the right way of doing this is?

    PS: Latex wouldn't work, so I left the code. If any moderator could fix this, I would be very grateful.
    Last edited by ymar; May 30th 2011 at 06:03 AM. Reason: PS
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by ymar View Post
    I have to calculate the Lebesgue measure of
    \left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}
    Now we can see the formula.
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  3. #3
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    Thank you, Fernando!

    PS I have just noticed that the title of my thread is misleading. Of course the set is in \mathbb{R}^3. I made a mistake.
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    Quote Originally Posted by ymar View Post
    I have to calculate the Lebesgue measure of

    \left\{(x,y,z)\,:\,x^2+y^2\leq 1,\,(1-x^2-y^2)^{k}z^2\leq(x^2-y^2)^2\right\}

    for every k\in\mathbb{Z}
    After the substitution I get \int_U r drd\alpha dz, where

    U=\left\{0<r<1,\,\alpha\in(0,2\pi),\,(1-r^2)^kz^2<r^4\cos^{\color{red}2}2\alpha\right\}. Note the square there!


    I can't calculate it. Everything I try leads to difficult integrals. Could you please give me an idea of what the right way of doing this is?
    The condition on z for a point to lie in U is -z_0<z<z_0, where z_0 = \frac{r^2|\cos2\alpha|}{(1-r^2)^{k/2}}. The integral for negative z is the same as for positive z, and so

    \int_U r\, dzd\alpha dr = 2\int_0^1\int_0^{2\pi}\int_0^{z_0} r\, dzd\alpha dr = 2\int_0^1\int_0^{2\pi}\frac{r^3|\cos2\alpha|}{(1-r^2)^{k/2}}d\alpha dr (doing the z integral first).

    The variables conveniently separate and give you 2\int_0^{2\pi}|\cos2\alpha|\,d\alpha\int_0^1\frac{  r^3}{(1-r^2)^{k/2}}\,dr. Check that \int_0^{2\pi}|\cos2\alpha|\,d\alpha = 8\int_0^{\pi/4}\cos2\alpha\,d\alpha = 4. For the other integral, substitute s=r^2 and get the integral in the form

    \tfrac12\int_0^1\frac s{(1-s)^{k/2}}\,ds = \tfrac12\int_0^1\frac {1-(1-s)}{(1-s)^{k/2}}\,ds = \tfrac12\int_0^1\bigl((1-s)^{-k/2} -(1-s)^{1-(k/2)}\bigr)\,ds,

    which you can easily integrate. I get the answer for the measure of U to be \frac{16}{(2-k)(4-k)} (but don't trust my accuracy there). Note that this becomes infinite when k=2. I guess that U must have infinite measure whenever  k\geqslant2.
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    Opalg, thank you. I would never notice the lacking square. I think the problem wouldn't be solvable without it, am I right?
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