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Math Help - Center of Mass

  1. #1
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    Feb 2009
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    Center of Mass

    Problem:

    A solid has the shape of the region bounded by the paraboloid z= x^2 + y^2 and the plane  z=4. The density at (x,y,z) is z - (x^2 + y^2). Find the center of mass of this solid.

    For double integral problems, the center of mass coordinates would be x= My/m and y= Mx/m (where Mx and My are moments of x and y, and m is mass), but I'm not sure how to go about what looks like to be a triple integral problem.

    To find the moments of x, y, and z, would you simply take the triple integral of each variable times the density z - (x^2 + y^2)?

    Appreciate any help. Thanks!
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  2. #2
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    May 2009
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    Hello wilcofan3! First thing I would do when dealing with any C.M. problem is get your head around the geometry of the problem. What I'm referring to is the fact that picturing the geometry can lead to simplifications in the calculation. Here you have a paraboloid which is rotationally symmetrical about the z-axis, as is its density. You can see this by transforming to cylindrical coordinates ( r,\theta, z) where the paraboloid surface becomes
    z = r^2
    and the density transforms to
    \rho = z -r^2.

    As you can see there are no \theta's in these equations meaning the problem is invariant to rotations about the z-axis. Hence, the C.M. must lie on the z-axis i.e. x=0 and y=0.

    Now, to compute the z-coordinate of the C.M. you need to work out the aggregate density function along the z-axis. What I mean by this is find the integral over r and \theta of the density function to give a function only in terms of z. This gives the effective density if the object was squashed into the z-axis. So integrating in cylindrical coordinates we have that

    \rho(z) = \int_0^{\sqrt{z}}\int_0^{2 \pi} (z-r^2 ) \, r \, \mathrm{d}\theta \, \mathrm{d}r = \frac{\pi}{2} z^2 .

    Then for the final part we're left with the usual one-dimensional moments type of calculation ( define \mu_z as z-coordinate of C.M.):

    \mu_z = \frac{\int_0^4 z \rho(z) \mathrm{d}z}{\int_0^4 \rho(z) \mathrm{d}z}.

    You should be able to take it from there.
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