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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:

[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. :)