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Math Help - Triple Integral Problem ><

  1. #1
    Newbie iSh4wty's Avatar
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    Triple Integral Problem ><

    Hello there!

    I have a question I've been stuck on for a couple days...would really appreciate some help! =) =) =)

    Find the volume of the region bounded by...
    the cone x^2 + y^2 = z^2 (z >= 0)
    and the paraboloid z = 2x^2 + 2y^2.

    Cheers in advance yo! ;D
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  2. #2
    Senior Member Spec's Avatar
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    V=\iint_D \left( \int_{2x^2+2y^2}^{\sqrt{x^2+y^2}}dz\right)dxdy where D is the projection on the xy-plane: \sqrt{x^2+y^2}=2x^2+2y^2

    Switch to polar coordinates and we get r=2r^2 \Longleftrightarrow r=0,\ r=\frac{1}{2}

    So we have D'=\left\{0\leq r\leq \frac{1}{2},\ 0\leq \varphi \leq 2\pi \right\} and the Jacobian is r.
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  3. #3
    Newbie iSh4wty's Avatar
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    Thanks for the speedy response! =D

    Another question: You've given me the limits for the third integral, I've found the limits for the second integral, how do I find them for the first? I try to put the two equations in terms of y and let them equal, but then I can't get rid of an x or z completely.

    Much appreciated. =)
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  4. #4
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    I've already given you all the limits you need. The area D is transformed to D' when switching to polar coordinates, so the integral we're left with is:

    \iint_D \left( \int_{2x^2+2y^2}^{\sqrt{x^2+y^2}}dz\right)dxdy=\ii  nt_D(\sqrt{x^2+y^2}-(2x^2+2y^2))dxdy=\iint_{D'}(r-2r^2)r\ drd\varphi
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