Calculating the Area of a Circle on S^2 in the Spherical Metric

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Homework Help Overview

The discussion revolves around calculating the area of a circle on the surface of a sphere (S^2) using the spherical metric. The original poster seeks to demonstrate that the area is given by the formula 2π(1 - cos p), where p is the radius of the circle.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss dividing the circular region into ring-shaped slices and integrating to find the area. There is uncertainty about the method of calculating the area of each ring and how to apply the spherical metric to determine the perimeter.

Discussion Status

Some participants have offered guidance on how to approach the problem by suggesting the division of the circle into rings and the integration of their areas. However, there remains a lack of clarity regarding the application of the metric and the calculations involved.

Contextual Notes

Participants express confusion about the initial instructions and the mathematical concepts involved, indicating a need for further clarification on the setup and definitions used in the problem.

halvizo1031
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I need help with this problem:

given a cirlce on S^2 of radius p in the spherical metric, show that its area is 2pi(1-cos p)
 
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Welcome to PF!

Hi halvizo1031! Welcome to PF! :smile:

(have a pi: π and a rho: ρ and try using the X2 tag just above the Reply box :wink:)
halvizo1031 said:
given a cirlce on S^2 of radius p in the spherical metric, show that its area is 2pi(1-cos p)

Divide the circular region into ring-shaped slices of thickness ds, and integrate …

what do you get? :smile:
 
I'm not sure I understand what you wrote.
 
halvizo1031 said:
I'm not sure I understand what you wrote.

Divide the circle into rings …

the area of each ring is its thickness times its length (ie its perimeter) …

use the metric to find the length of the perimeter of each ring …

then add up the areas of all the rings :smile:
 
ok I'll give that a try. thanks!
 

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