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Volume of sphere

  1. Oct 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the theorem of pappus to find the volume of the given solid
    A sphere of radius r
    2. Relevant equations

    V = 2∏xA



    3. The attempt at a solution

    V = 2∏(4r/3∏)(4∏r^2) = (16/3)∏r^3

    So something is wrong I should end up with (4/3)∏r^3 no?
     
  2. jcsd
  3. Oct 17, 2013 #2

    Dick

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    I don't think this makes any sense at all, which may be why you got the wrong answer. Which theorem of Pappus are you using? What surface are you revolving to get the sphere, and where is it's centroid?
     
  4. Oct 17, 2013 #3
    I used the surface area of a sphere. 4PIr^2 and then the (4r/3)PI is centroid of semi circle from the example they told me to use in my book.

    Should I use a circle ? V = 2PI(4r/3PI)(PIr^2) still doesn't make sense though.
     
  5. Oct 17, 2013 #4

    SteamKing

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    What's the area of a semicircle of radius R?
     
  6. Oct 17, 2013 #5

    Dick

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    No, still doesn't. You still haven't stated coherently what you are doing. What theorem of Pappus and what are the parts?
     
  7. Oct 18, 2013 #6
    Theorem of Pappus


    Let R be a plane region that lies entirely on one side of a line l in the plane. If R is rotated about l, then the volume of the resulting solid is the product of the area A of R and the distance d travled by the centroid of R.

    Question: Use the theorem of Pappus to find the Volume of the given solid.

    44.) A sphere of radius r (use example 4)

    Example 4 is Find the center of mass of a semicircular plate of radius r.

    The result is: The center of mass is located at the point 90, 4r/(3PI))


    OK so the question I'm owrking in is in bold.

    Can you help me now is this enough information? Should I not end up with the volume of a sphere V = (4/3)∏r^3 ???
     
  8. Oct 18, 2013 #7
    What are you using for the area of the semicircle?
    The plane you are rotating is a semicircle, so you need to use its area (the area of a semicircle).
     
  9. Oct 18, 2013 #8
    Should I use (1/2)PIr^2 ?
     
  10. Oct 18, 2013 #9
    Yes :) That would be the area of a semicircle, and should give you the right result.
     
  11. Oct 18, 2013 #10

    SteamKing

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    If jbreezy would check his work and the suggestions in posts #4 and #7, the answer would fall out immediately.
     
  12. Oct 18, 2013 #11
    I'm sorry this doesn't make any sense to me. So I have the area of the semi circle that is A in the equation
    V = 2PIxA where x is the distance travled by the centroid? So I would have V = 2PI(PIr^2)((1/2)PIr^2 )
    I used PIr^2 for x.
     
  13. Oct 18, 2013 #12

    Dick

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    You are garbling the theorem. The theorem says V=Ad, where d is the distance travelled by the centroid. Since the centroid travels in a circle if the radius of that circle is x then the distance d=2*pi*x. So x is the RADIUS of the circle the centroid makes, not the distance travelled by the centroid. So what is x?
     
  14. Oct 18, 2013 #13
    I'm not garbling anything. I'm giving you the exact words from my book they are the ones who write this trash.

    I did this

    V = Ad = 2PI(4r/3PI)(PIr^2/2)

    A = (PIr^2/2)
    d = 2PI((4r/3PI))
    Is this right?
     
  15. Oct 18, 2013 #14

    Dick

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    Yes, but you should write the centroid radius more carefully to make it clear the pi is in the denominator. It's 4r/(3pi). Doesn't V come out to be what you expect?
     
    Last edited: Oct 18, 2013
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