Is One Concentric Circle of a Platter, or the Surface Area That Passes Under a Read/write Head

AmagicalFishy

Hey, folks.

I'g trying to derive the surface 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 whatsoever such thing, I want to work geometrically direct from spherical coordinates. I am trying to do this by integrating concentric rings. Hither's a picture of what I'm talking about:

[tex]\phi - \text{ is the Azimuth (annotation; there is one instance at the middle and one about the elevation for illustration)}\\
\theta - \text{ is the Zenith}\\
r - \text{ is the radius of the circumvolve currently beingness integrated} \\
R - \text{ is the radius of the sphere}[/tex]

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

In a sphere, the radius r of the integrated-circles varies co-ordinate to the Zenith. The radius is:
[tex]r = R sin{\theta}[/tex]
Then, the circumference of any given circumvolve within the sphere, at a pinnacle designated by θ, is:
[tex]\int_0^{2\pi}R sin{\theta} \ \mathrm{d}\phi[/tex]
That is, the radius of a circumvolve multiplied past 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, and then θ has to move from top-to-bottom. The limits of integration on θ are therefore from zippo 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: four[itex]\pi[/itex]R
... which is shut, but wrong. If I threw in another R, life would be expert, simply 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 in a higher place makes pretty intuitive sense to me, and I tin can't point out where or why I would add together in some other R. Though it's often the case that I'm making a dumb mistake.

Thank you. :)

marcusl

To stay in spherical coordinates, you need to write the differential element of expanse in spherical coordinates. In the [itex]\chapeau\phi[/itex] direction, the differential arc is [itex]rd\phi[/itex]. In the [itex]\chapeau\theta[/itex] direction, the differential arc is [itex]r\sin\theta d\theta[/itex], as y'all tin can convince yourself by drawing a diagram or looking in a calculus volume. Thus the differential area is [itex]r^ii\sin\theta d\theta d\phi[/itex] , and the total surface area is [tex]r^2\int_0^{ii\pi}d\phi\int_0^\pi \sin\theta d\theta.[/tex]

AmagicalFishy

Await.

Oh, wait, I get information technology!

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

Give thanks you. :)

Vargo

Here is a cool fact that is related to your question.

Wrap a cylinder around that sphere. i.e. x^ii+y^2 = R^ii with z between -R and R. Any region on the sphere has the aforementioned surface area every bit the respective area on the cylinder. The correspondence is via a radial projection out from the z axis. So, for example, the area between latitudes would exist 2pi*R^2(cos(phi1)-cos(phi2)).

olivermsun

Hither is a cool fact that is related to your question.
Wrap a cylinder effectually that sphere. i.eastward. ten^2+y^ii = R^2 with z between -R and R. Any region on the sphere has the same area every bit the corresponding area on the cylinder.

That is pretty absurd!

marcusl

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

Watley Weepleget

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