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Thread: Tripple integration question spherical or cylindrical coordinates

  1. #1
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    Tripple integration question spherical or cylindrical coordinates

    Use cylindrical or spherical coordinates, whichever seem more appropriate,
    to evaluate the triple integral $\displaystyle \int \int \int_{V} z dV $ where V lies above the paraboloid $\displaystyle z = x^2 + y^2$ and below the plane $\displaystyle z = 6y.$

    I am having trouble with the bounds and am not sure whether to use cylindrical or spherical coordinates.
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  2. #2
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    Re: Tripple integration question spherical or cylindrical coordinates

    I guess I can see it in cylindrical:

    z is going...

    from r^2 (the paraboloid) up to 6y i.e. 6r sin theta (the plane roof)

    ... for every bit that r is going...

    from zero (the z axis) along to as far as where r^2 = 6r sin theta => r = 6 sin theta

    ... for every bit that theta is going from zero (the +ive x axis) round to pi (-ive). So,

    $\displaystyle \displaystyle{V = \int_{0}^{pi}\ \int_{0}^{6 \sin \theta} \int_{r^2}^{6r \sin \theta} rz\ dz\ dr\ d\theta}$

    int from 0 to pi int from 0 to 6sin theta int from r^2 to 6rsin theta rz dz dr dtheta - Wolfram|Alpha

    The integral graph there (going 0 to pi on the horizontal) looks about right.

    Those zero sins aren't going to work, but see also,

    integrate 1944 sin^6 theta - Wolfram|Alpha

    which gives 121.5 pi. If this isn't wrong I'll put a pic for the computation similar to

    http://www.mathhelpforum.com/math-he...tml#post621761

    when I'm home, could be a while.



    Anyway, hope that helps.


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    Last edited by tom@ballooncalculus; Sep 8th 2011 at 04:53 PM. Reason: 121.5
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  3. #3
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    Re: Tripple integration question spherical or cylindrical coordinates

    Yeah that looks good to me. Thanks! That last integral is nasty.
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