Surface area of a sphere-derivation

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SUMMARY

The surface area of a sphere can be derived using double integration by correctly applying spherical coordinates. The differential area element is given by r2sin(θ) dθ dφ, not r2 dθ dφ. Integrating this expression over the limits (0, 2π) for φ and (0, π) for θ yields the correct surface area of 4πr2. The initial approach was flawed due to an incorrect differential area element.

PREREQUISITES
  • Spherical coordinates
  • Double integration techniques
  • Understanding of polar and azimuthal angles
  • Basic calculus
NEXT STEPS
  • Study the derivation of surface area in spherical coordinates
  • Learn about the application of double integrals in various geometries
  • Explore the concept of differential area elements in calculus
  • Review the relationship between polar coordinates and spherical coordinates
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Mathematicians, physics students, and anyone interested in geometric derivations and calculus applications.

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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