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Finding the Centroid of a Solid

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the volume and the centroid of the solid E that lies above the cone z=√x^2+y^2 and below the sphere x^2+y^2+z^2=49.

    2. Relevant equations


    3. The attempt at a solution

    My bounds were:
    [itex]\theta[/itex]=0 to 2[itex]\pi[/itex]
    [itex]\varphi[/itex]=0 to [itex]\pi[/itex]/4
    [itex]\rho[/itex]=0 to 7cos([itex]\varphi[/itex])

    So my integral was:
    ∫∫∫p^2sin([itex]\varphi[/itex]) d(rho) d[itex]\varphi[/itex]) d(θ)

    I just need help in if my bounds and integral are correct. If someone can help me with that, that would be great!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2011 #2

    LCKurtz

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    That last limit is wrong. What is the spherical coordinate equation of the sphere [itex]\rho=?[/itex]

    And of course, once you correct the limits, that formula just gives you the volume. I think you undertand there is more to do, right?
     
  4. Oct 30, 2011 #3
    So is ρ from 0 to cos(φ)? I'm still confused about this.

    And yes I understand that I still have to solve the integrals :)
     
  5. Oct 30, 2011 #4

    LCKurtz

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    What is the equation of the sphere in spherical coordinates? Your integral must go from ρ = 0 to ρ on the sphere. So you need the equation of the sphere in spherical coordinates. Hint: It is very simple!
     
  6. Oct 30, 2011 #5
    So the equation of the sphere in spherical coordinates is ρ^2=x^2+y^2+z^2. So the integral should go from p=0 to 7, right?
     
  7. Oct 30, 2011 #6

    LCKurtz

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    Yes! But actually the equation of the sphere in spherical coordinates is ρ = 7, not ρ = x2+y2+z2.
     
  8. Oct 30, 2011 #7
    Ah! Ok I got it now, thank you so much for helping!
     
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