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Small question

  1. May 18, 2004 #1
    If I intergrate 2(pi)r I get (pi)r^2

    If I integrate that I get 1/3 (pi)r^3, which is close to the volume of a sphere. But where do I get a 4/3(pi)r^3 ?
     
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  3. May 18, 2004 #2

    arildno

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    Differentiate the volume expression, and find a suitable interpretation of the result.
     
    Last edited: May 18, 2004
  4. May 18, 2004 #3

    matt grime

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    beacause disks are not the infinitesimal shells that add up to a solid sphere
     
  5. May 18, 2004 #4
    Would it work if I used half of the area forumula, and used the disk method to rotate it about an axis? Should that give me 4/3(pi)r^3?

    I would find out for myself but I'm a little busy.

    Thanks for the replies.
     
  6. May 18, 2004 #5

    matt grime

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    that's a different technique: add up the shells of surface area of a sphere
     
  7. May 19, 2004 #6
    OKay I'm starting to understand.

    So I use the surface area formula(for a sphere). Integrate from 0 to r? :uhh:

    Edit: Oh my god. I feel so stupid...

    You see, I didn't know the formula for the surface area of a sphere. ......

    Oh well. Anyways, uum. So then, what is 1/3(pi)r^3? Without looking it up, making a wild guess...is it the volume of a cylinder?

    Edit2: I really need to stop thinking.... I have no clue what I'm talking about. :yuck: So is 1/3(pi)r^3 just nonsense?

    Edit3: Interesting that a cone's volume is 1/3(pi)r^2 h. Good I have much time to think about it.
     
    Last edited: May 19, 2004
  8. May 19, 2004 #7

    matt grime

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    the surface area of a sphere is 4(pi)r^2
     
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