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Surface area of a sphere

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the area for 3D sphere.

    2. Relevant equations
    I know there's this formula for surface of revolution:
    [tex]A=2\pi\int_{a}^{b}f(x)\sqrt{1+ f'(x)^2}\:\mathrm{d}x[/tex]

    3. The attempt at a solution
    I thought of dividing the the sphere into slices, each of which contains a ring.
    The length of each ring is [itex]2\cdot\pi\cdot r[/itex], with [itex]r=\sqrt{R^2-x^2}[/itex].
    We could then integrate:
    [tex]\int_{-R}^{R}2\pi\sqrt{R^2-x^2}\:\mathrm{d}x=4\pi\int_{0}^{R}\sqrt{R^2-x^2}\:\mathrm{d}x=\pi R^2[/tex]
    But this is not correct so there must be something wrong...

    PS: Just out of curiosity, is there any way to prove the formula for the surface are of an n-sphere using calculus? (the one with Γ)
     
  2. jcsd
  3. Mar 17, 2009 #2
    The integral which you computed is (obviously) for area based on your answer. This is because you're taking a whole bunch of rings with an infinitely small width and summing them up from 0 to R. Geometrically think of it as taking a ring and fitting successively smaller rings inside of it until the point at which all the rings together resemble a single solid. That is why you are computing the area instead of surface area.

    http://en.wikipedia.org/wiki/Hypersphere see the portion on volume.
     
    Last edited: Mar 17, 2009
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