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Thread: triple integral and center of mass

  1. #1
    Junior Member
    Joined
    Jun 2009
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    triple integral and center of mass

    hi there,

    my question is:

    i have a D = {(x,z): ax^2 <= z <= h)}, where a and h are parameters.

    if you take this D and rotate it around the z axis,

    you get a shape in r^3 - which we'll call V.

    we start changing the a parameter, so V changes.

    how do i need to change h so that the z coordinate of the center of

    mass remains in the same point (the body's mass density is constant)?


    i realized that the shape of V is a paraboloid whose outer "radius" is sqrt(h/a) and runs on z from 0 to h. i've tried to calculate a
    triple integral of z*dx*dy*dz / the triple integral of 1*dx*dy*dz, but get the answer wrong.

    i know my description of the question is somewhat long,
    but please help, and thanks in advance...
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  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    The lower boundary of the volume of revolution $\displaystyle V$ of $\displaystyle z=ax^2$ about the $\displaystyle z$-axis is given by

    $\displaystyle z=ar^2\quad\quad 0\le z \le h.$

    You are correct that to find the $\displaystyle z$-coordinate of the center of mass of $\displaystyle V$, $\displaystyle z_M$, we divide

    $\displaystyle \int\int\int z\,dx\,dy\,dz\quad\quad\mbox{by}\quad\quad\int\int \int\,dx\,dy\,dz.$

    To find each, we may convert to cylindrical coordinates:

    $\displaystyle \int_0^{2\pi}\int_0^{\sqrt{\frac{h}{a}}}\int_{ar^2 }^h z\,dz\,dr\,d\theta,$

    or, alternatively, use the Disc Method:

    $\displaystyle \int_0^h \pi\left(\sqrt{\frac{z}{a}}\right)^2z\,dz.$

    I found the same answer for each, but it did not depend on the value of $\displaystyle a$.
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