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Surface Area of a Sphere in Spherical Coordinates; Concentric Rings

  1. Mar 7, 2013 #1
    Hey, folks.

    I'm trying to derive the surface area of a sphere using only spherical coordinates—that is, starting from spherical coordinates and ending in spherical coordinates; I don't want to convert Cartesian coordinates to spherical ones or any such thing, I want to work geometrically straight from spherical coordinates. I am trying to do this by integrating concentric rings. Here's a picture of what I'm talking about:

    Sphere_zps94366c33.jpg

    [tex]\phi - \text{ is the Azimuth (note; there is one instance at the center and one near the top for illustration)}\\
    \theta - \text{ is the Zenith}\\
    r - \text{ is the radius of the circle currently being integrated} \\
    R - \text{ is the radius of the sphere}[/tex]

    I began simply by deriving the equation for the circumference of any circle:
    [tex]\int_0^{2\pi}\!r\ \mathrm{d}\phi[/tex]
    (The arc-length is the circle's radius multiplied by the angle; d[itex]\phi[/itex] is the infinitesimal angle, so the integrand is the infinitesimal arc-length.)

    In a sphere, the radius r of the integrated-circles varies according to the Zenith. The radius is:
    [tex]r = R sin{\theta}[/tex]
    Then, the circumference of any given circle within the sphere, at a height designated by θ, is:
    [tex]\int_0^{2\pi}R sin{\theta} \ \mathrm{d}\phi[/tex]
    That is, the radius of a circle multiplied by the infinitesimal angle, from zero to two-pi, as shown above, will give you the circumference of that circle.
    Now, we want the integral to sum all of the circles' circumferences of the sphere, so θ has to move from top-to-bottom. The limits of integration on θ are therefore from zero to pi, and the whole integral is:
    [tex]\int_0^\pi \int_0^{2\pi}\! \!R sin{\theta} \ \ \mathrm{d}\phi \ \mathrm{d}\theta[/tex]

    This evaluates to: 4[itex]\pi[/itex]R
    ... which is close, but wrong. If I threw in another R, life would be good, but I can't well do that without knowing why. I've thought of why I should need another R, but I can't figure anything out.

    Where am I missing an R factor, and why should I be factoring it in? Everything above makes pretty intuitive sense to me, and I can't point out where or why I would add in another R. Though it's often the case that I'm making a dumb mistake.

    Thanks. :)
     
  2. jcsd
  3. Mar 7, 2013 #2

    marcusl

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    To stay in spherical coordinates, you need to write the differential element of area in spherical coordinates. In the [itex]\hat\phi[/itex] direction, the differential arc is [itex]rd\phi[/itex]. In the [itex]\hat\theta[/itex] direction, the differential arc is [itex]r\sin\theta d\theta[/itex], as you can convince yourself by drawing a diagram or looking in a calculus book. Thus the differential area is [itex]r^2\sin\theta d\theta d\phi[/itex] , and the total surface area is [tex]r^2\int_0^{2\pi}d\phi\int_0^\pi \sin\theta d\theta.[/tex]
     
  4. Mar 8, 2013 #3
    Wait.

    Oh, wait, I get it!

    Haha! Excellent. At one point, I was on the verge of that—then I realized that the radius of the circle being integrated varied with the sin of the zenith, and (I don't know why) decided on concentric rings.

    Thank you. :)
     
    Last edited: Mar 8, 2013
  5. Mar 8, 2013 #4
    Here is a cool fact that is related to your question.

    Wrap a cylinder around that sphere. i.e. x^2+y^2 = R^2 with z between -R and R. Any region on the sphere has the same area as the corresponding area on the cylinder. The correspondence is via a radial projection out from the z axis. So, for example, the area between latitudes would be 2pi*R^2(cos(phi1)-cos(phi2)).
     
  6. Mar 8, 2013 #5

    olivermsun

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    That is pretty cool!
     
  7. Mar 8, 2013 #6

    marcusl

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    That was actually proven by Archimedes and published in 225 BC. He was a smart guy...
     
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