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Find the volume and centroid of the solid

  1. Jul 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the volume and centroid of the solid E that lies above the cone z= sqrt (x^2 + y^2) and below the sphere x^2 + y^2 + z^2 = 1.

    3. The attempt at a solution

    I found the correct volume=(pi/3)(2-sqrt2)

    How do I find the centers of mass if I don't know the mass and can't find the mass since I have no density function?
  2. jcsd
  3. Jul 18, 2008 #2
    It asks for "centroid" not center of mass. The difference is that the centroid is is the center of gravity in the special case of a homogeneous object, meaning that density is constant so it cancels out.

    [tex]x_{centroid} = \frac{\int\int\int x dV}{V}[/tex]
  4. Jul 18, 2008 #3
    Is it necessary to divide by the volume then? If the density is constant, wouldn't the mass and the volumes cancel out? So we would just be left with the triple integral of x dV.
  5. Jul 18, 2008 #4
    What do you mean by "wouldn't the mass and the volumes cancel out"?

    If it helps, I can show where the canceling specifically occurs:

    x_{center of mass} = \frac{\int\int\int x \sigma(x,y,z) dV}{M}[/tex]

    Then use the fact that [tex]\sigma = dM/dV[/tex] so [tex] M = \int\int\int \sigma(x,y,z) V[/tex]

    so [tex]x_{center of mass} = \frac{\int\int\int x \sigma(x,y,z) dV}{\int\int\int \sigma(x,y,z) V}[/tex]

    [tex]\sigma(x,y,z)[/tex] is the density function. It is constant so it can be taken out of the integrand.

    x_{center of mass} = \frac{\sigma(x,y,z) \int\int\int x dV}{\sigma(x,y,z)\int\int\int V}[/tex]

    And that's where it cancels from, so you are left with the expression above for homogeneous density.
  6. Jul 19, 2008 #5


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    Staff Emeritus
    Science Advisor

    There is no "mass' and there is no "center of mass". Your problem does not ask for "center of mass", it asks for "centroid". What is the definition of "centroid"?
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