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Homework Help: Find the center of mass with tripl-integral

  1. Jan 10, 2010 #1
    A being is on a sphere like planet. Sudently hey grabs a GPS and starts diging straight on a diemeter, heading to the center. We can theoretically say tha he is carrying a level with him. Then tha sphere is theoretically cut in two section , A and B. I have(well...with help) found the volume of the two sections with a tripl-integral using Spherical coordinate system.We know that the sphere has the same density all over (an inside) the sphere. I neeeeeeeeeeeeed to find the center of mass. Odviously the center of mass is going to be on the z Axle. Can someone help me find the center of mass?
     

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  3. Jan 10, 2010 #2

    ideasrule

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    Homework Helper

    If you can find the mass, why can't you find the center of mass? It's just the integral of rho*dV*z whereas the mass is the integral of rho*dV*z.
     
  4. Jan 10, 2010 #3
    I don't quite understand what the question is about, with the digging and stuff o.0
    For a sphere with uniform density, clearly the centre of mass must be at the centre of the sphere.
     
  5. Jan 10, 2010 #4
    Well what i ment is that i need to find the center of mass of each section , A and then B.
     
  6. Jan 10, 2010 #5
    ideasrule, i fail to see the difrens betwen the two integral, also rho=?
     
  7. Jan 10, 2010 #6
    Oh ok I get it now. For each region just apply the usual formulas with the appropriate parameters:
    [tex]
    \overline{x} = \frac{1}{M} \iiint x \rho (x,y,z) dV
    [/tex]
    [tex]
    \overline{y} = \frac{1}{M} \iiint y \rho (x,y,z) dV
    [/tex]
    [tex]
    \overline{z} = \frac{1}{M} \iiint z \rho (x,y,z) dV
    [/tex]
     
  8. Jan 10, 2010 #7
    I can understand that i am only going to use the z formula. I alsow understand it is going to be a tripl-integral. What i cant find is from where to where am i going to integrat, and what coordinate system would be most suted
     
  9. Jan 10, 2010 #8
    Since it is a sphere, definitely we will be using spherical coordinates:
    [tex]
    \psi (\rho, \theta, \phi) = (\rho\,sin \phi\, cos \theta, \rho\, sin \phi\, sin \theta, \rho\, cos \phi)
    [/tex]
    for both regions A and B.
    However, the parameters have different ranges.
    For A:
    [tex]0 \leq \rho \leq R\,\, , 0\leq \theta \leq 2\pi\,\, , 0\leq \phi \leq a[/tex]
    For B:
    [tex]0 \leq \rho \leq R\,\, , 0\leq \theta \leq 2\pi\,\, , a\leq \phi \leq \pi[/tex]
    I presume you are able to convert the triple integral in rectangular coordinates into spherical coordinates?

    --Scrap this, i made a terrible mistake--
     
    Last edited: Jan 10, 2010
  10. Jan 10, 2010 #9
    you meen dV=dz*dy*dx=r^2*sinθ*dr*dθ*dφ ?? Also by a you meen the corner acording to the depth?
     
  11. Jan 10, 2010 #10
    Yup. And remember to convert z as well.
    Wait I think I made a blunder in my parameters...oops I did a cone instead...gimme a while to fix it
     
  12. Jan 10, 2010 #11
    "Yup. And remember to convert z as well" what do you meen?
     
  13. Jan 10, 2010 #12
    Scrap those, we'll return to good old rectangular coordinates for now. If I'm not mistaken, the region A can be represented as
    [tex]\int_{-\sqrt{R^2-a^2}}^{\sqrt{R^2-a^2}}\,\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\,\int_{a}^{\sqrt{R-x^2-y^2}}\,\,dz\,dy\,dx[/tex]
    where a = height of base of region above the origin.
    Looks pretty ugly though, so maybe someone else wants to help out here?
     
  14. Jan 10, 2010 #13
    if this helps, when calculating the volume i had to solve this:
    triple-integral{(r from b/Cosθ to R,θ from 0 to ArcCos[b/R],φ from 0 to 2π)r^2 sinθ dr dθ dφ} R = radius, b = R - depth of being
     
  15. Jan 10, 2010 #14
    Sadly, my calculus remains rather weak.
    Ah..So those will be the limits of the centre of mass integral as well. Just substitute z with the parameterised equivalent.
     
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