Surface Area of a Sphere in Spherical Coordinates; Concentric Rings

[AmagicalFishy]

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:

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

I began simply by deriving the equation for the circumference of any circle:
\int_0^{2\pi}\!r\ \mathrm{d}\phi
(The arc-length is the circle's radius multiplied by the angle; d\phi 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:
r = R sin{\theta}
Then, the circumference of any given circle within the sphere, at a height designated by θ, is:
\int_0^{2\pi}R sin{\theta} \ \mathrm{d}\phi
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:
\int_0^\pi \int_0^{2\pi}\! \!R sin{\theta} \ \ \mathrm{d}\phi \ \mathrm{d}\theta

This evaluates to: 4\piR
... 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. :)

 

[marcusl]

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

 

[AmagicalFishy]

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. :)

 

[Vargo]

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)).

 

[olivermsun]

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.

That is pretty cool!

 

[marcusl]

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