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Finding centroid of a solid

  1. Jul 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Use cylindrical coordinates to find the centroid of the solid.

    The solid that is bounded above by the sphere x2 + y2 + z2 = 2

    and below by z = x2 + y2

    2. Relevant equations

    x = rcos(theta)

    y= rsin(theta)



    3. The attempt at a solution

    I am having trouble trying to find the limits of integration for r.

    I have been able to get the sphere in terms of r as z = (2-r2)1/2 and the paraboloid as z = r2

    I understand that in cylindrical coordinates the region is occupied from 0<theta<2pi and r originates from origin n to the uppermost part of the sphere.

    I also found that when you equate both surface z equations. you get r2 = (2-r2)1/2 . Solving for r you get radical 2 and 1. r being the furthest distance would then have the upper limit of 1. I'm not sure if this is the correct method of solving.
     

    Attached Files:

  2. jcsd
  3. Jul 22, 2015 #2

    RUber

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    First off, ## \sqrt 2>1## and ##\sqrt 2## doesn't seem to solve your equation.

    Go with z=1 for the intersection point, so z is defined by the sphere above 1<z <2 and as the paraboloid for 0<z <1.
     
  4. Jul 22, 2015 #3

    RUber

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    See the attached image, this is the shape you are dealing with (revolved around the z axis to form a volume). Clearly, your centroid will be at x=0, y=0. How will you find the centroid in the z direction?
    PF_1.jpg
     
  5. Jul 22, 2015 #4
    So what I did was found out the value of z at the intersection point by equating both surfaces in terms of r, which led z to be either -2 or 1. When inputting z =1 i get r to also be 1

    The centroid in the z direction will thus be 02pi 01 r2√(2-r2) z ⋅ r dz dr dθ all divided by the volume of the region
     
    Last edited: Jul 22, 2015
  6. Jul 22, 2015 #5

    RUber

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    Do you mean you solved ##r^2 = (2-r^2)^{1/2}## as a quadratic to find ## r^2 = 1 \text{ or } -2 ## and then discarded the imaginary solution?
    Any troubles with these integrals?
     
  7. Jul 22, 2015 #6
    Yes I discarded the imaginary solution, I have no problem solving the integrals. I only had trouble identifying the upper limit of r. Would you say that the way I solved for r was a correct solution?
     
  8. Jul 22, 2015 #7

    RUber

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    That is the correct way to do it.
    What you did wrong on your original work posted in the image was to go from
    ## r^4 + r^2 =2 ## to r =1 or r = sqrt(2); you should have gotten r =1 or r = sqrt(-2).
     
  9. Jul 22, 2015 #8
    Ah I overlooked the proper solution for r. Thanks for taking the time to help me out!
     
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