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Surface Area of Sphere

  1. Jan 8, 2008 #1
    The given surface area of a sphere is 4*(pi)*r^2.
    There are several proofs to this, but I'm just looking for the error in this one:

    For the arc length, s = rx, x being the angle in radians.
    Therefore, s = 2(pi)r = C, C being circumference of circle.
    The surface are of a sphere is the sum of the circumferences of a number of circles surrounding around a point from which the radius extends. Therefore, a set of n points is needed for n circles.
    For a sphere, if just a horizontal cross section was taken (that includes the center), it would come out as a horizontal circle of circumference 2(pi)r. If for each point on that circle, there was a vertical circle that surrounded the center, then only half the total number of points would be needed for the surface area of the sphere (since the other half would produce circles that would coincide with the first half). Thereforce, only half the length of the circle would be needed, being only (pi)r, which would be the value for n.
    Therefore, n number of points can be given by s = (pi)r. With (pi)r number of points, the sum of the circumferences of each of these points can be given by [(pi)r] * [2(pi)r], which results in 2*(pi)^2*r^2. The difference between this and the official formula is by and additional coefficient of (pi/2). I'm having trouble spotting the error in this proof.
     
  2. jcsd
  3. Jan 8, 2008 #2

    Dick

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    That is so distorted I don't know where to start. How about "The surface area of a sphere is the sum of the circumferences of a number of circles surrounding around a point from which the radius extends." I don't think I ever saw that in Euclid. Do you believe it?
     
  4. Jan 8, 2008 #3
    Cant exactly tell. Whats wrong with the concept?
     
  5. Jan 9, 2008 #4

    Dick

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    For one thing the sum of lengths is a length, not an area. It's dimensionally wrong. And that's only the beginning of it's utter wrongness.
     
  6. Jan 9, 2008 #5
    ಠ_ಠ....I see that you are right.
     
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