Surface area of a sphere-derivation

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Homework Help Overview

The discussion revolves around deriving the surface area of a sphere using double integration, specifically in the context of a previous electromagnetism problem. The original poster explores the concept of dividing the sphere's surface into infinitesimal regions to compute the area.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the surface area by using polar coordinates and considers the area of infinitesimal regions. They express concern about being off by a factor in their integral setup.

Discussion Status

Some participants have provided feedback on the original poster's approach, suggesting a correction to the surface area differential. The conversation appears to be ongoing, with participants engaging in clarifying the mathematical representation involved.

Contextual Notes

The original poster mentions that this derivation is not a homework question but rather a lingering curiosity from a past problem. There is an indication of uncertainty regarding the correct formulation of the surface area differential in spherical coordinates.

muppet
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[SOLVED] Surface area of a sphere-derivation

This isn't really a homework question, it just would've been handy to be able to do for an electromagnetism problem last year, and has been bugging me since!
Is it possible to derive the surface area of a sphere by double integration?
At the time I tried diving the surface into many infinitesmal regions that could be considered approximately plane rectangles. Each of these regions had sides of length rd(phi) and rd(theta), where phi and theta are the polar and azimuthal angles. The area of these regions was therefore r^{2}d\theta d\phi
Computing the integral over the limits (0, 2\pi),(0,\pi)
you're out by a factor of 2/\pi.
Any suggestions?
 
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I figured it would be a correction for the tapering with height, but couldn't work out how- cheers :smile:
 
You're welcome.
 

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