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Derivation of the area of a sphere formula.

  1. Mar 12, 2005 #1
    Deduce the formula for the area of a sphere with ratio R. (I already know it is 4*pi*R^2)
     
  2. jcsd
  3. Mar 12, 2005 #2
    The volume of a ball is [tex]\frac{4}{3}{\pi}r^3[/tex] and the surface area of a ball is [tex]4{\pi}r^2[/tex]. The surface area is the derivative of the volume.

    Jameson
     
  4. Mar 12, 2005 #3

    dextercioby

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    Not really.The cube...:wink:

    The simplest is:
    [tex] S=\iint dS=R^{2}\int_{0}^{2\pi}d\varphi \int_{0}^{\pi} d\vartheta \ \sin\vartheta [/tex]

    Daniel.
     
  5. Mar 12, 2005 #4

    dextercioby

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    How do you prove the volume of the ball is the one that u mentioned & how do you prove that the area of the 2-sphere is the derivative (wrt radius) of the volume of the 3-ball...?

    Daniel.
     
  6. Mar 12, 2005 #5
    Yes, I have to prove that the volume of a sphere is [tex]\frac{4}{3}{\pi}r^3[/tex], and that the surface area of a sphere is [tex]4{\pi}r^2[/tex], without using integrals.
     
  7. Mar 12, 2005 #6

    dextercioby

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    The volume of a sphere is zero...There's a giant thread here (i think in the "Calculus & Analysis" forum) bearing this meaningless name (the volume of a sphere),in which Saltydog,Mathwonk & Galileo give Archimede's rationale...

    Daniel.
     
  8. Mar 17, 2005 #7
    The volume of a sphere can be obtained without using integrals by remarking that the volume of a cylinder of radius r and height r is equal with the volume of a cone having the radius of the base r and the height r + the volume of a semisphere of radius r. If I remember well the demonstration involved the use of the principle of Cavalieri. So the volume of a semisphere is (2/3)*π*r3. But I do not think that the area of a sphere can be obtained without calculus, at least I do not think there is an easy (and exact) method.
     
    Last edited: Mar 17, 2005
  9. Mar 17, 2005 #8

    uart

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    I dont know of any method that doesn't rely on caculus, but the volume is a pretty easy "volume of rotation" integration. Just set up a hemi-sphere as the volume of rotation of the function y=+sqrt(r^2 - x^2) over x=[0..r].

    dV = A dr can be obtained "by inspection".
     
  10. Mar 17, 2005 #9
    missing calculus....integrals are good to solve such.
     
  11. Mar 22, 2005 #10

    Alkatran

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    Start from the equation y = sqr(x^2 - r) (r is a constant)
    Integrate it to get the surface of a circle, pi*r^2
    Then say you are putting a whole bunch of discs, with volume pi*r^2 * dx along the x axis with radius y = sqr(x^2 - r) to get this function:

    Integral[-1 to 1]: sqr(x^2 - r)^2 * pi = pi*Integral[-1 to 1]: x^2 - r


    Of course, you'd also need the proofs that a semi circle is defined by y = sqr(x^2 - r) and all the calculus ones.
     
  12. Mar 23, 2005 #11

    Galileo

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    Archimedes solved the problem of finding the area of a sphere with given radius.

    I don't know how he did it, but I know it his solution counts as one of the most notable mathematical achievements.
     
  13. Mar 23, 2005 #12

    Alkatran

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    Whoops, my mistake, I meant to use the perimeter of a circle formula (2pi*r) * dx, which would give you the sum of the areas of the rounded edge of all the discs (which would approach sphere's area as dx -> 0)
     
  14. Feb 17, 2009 #13
  15. Jun 1, 2011 #14
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