Find the Surface Area of a Sector of a Sphere: Trig Equations

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Discussion Overview

The discussion centers around finding the surface area of a sector of a sphere, with a focus on trigonometric equations and their applications in physics. Participants explore various methods and formulas, including the concept of solid angles and their relevance to the problem at hand.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a list of trigonometric equations, specifically for the surface area of a sector of a sphere.
  • Another participant suggests using the formula s = r*θ, where θ is the angle of the sector.
  • A different participant recalls that the area of a sphere is 4πr² and proposes calculating the area of a sector as a fraction of the sphere's surface.
  • Concerns are raised about measuring angles on a sphere, emphasizing the need for solid angles rather than circular formulas.
  • A participant mentions needing to calculate light hitting an aperture from stars, indicating a practical application of the surface area calculation.
  • One participant provides a formula for the area of a sphere in relation to θ and suggests it should be in radians.
  • Another participant discusses the concept of solid angles and their representation as Ω, noting their limited exposure to the term in math courses.
  • There is a mention of the connection between solid angles and astrophysics, with curiosity about their relevance to mathematics.
  • Participants share experiences regarding the difficulty of courses like Electricity and Magnetism and their study habits.
  • Further questions arise about the relationship between arccot(x) and the slope of rays hitting a parabolic surface.

Areas of Agreement / Disagreement

Participants express varying views on the appropriate formulas and concepts for calculating the surface area of a sector of a sphere. There is no consensus on a single method or formula, and the discussion remains unresolved regarding the best approach.

Contextual Notes

Some participants reference the need for solid angles and the distinction between circular and spherical geometry, indicating potential limitations in understanding the problem without these concepts. There are also unresolved mathematical steps and dependencies on definitions that could affect the discussion.

Who May Find This Useful

This discussion may be of interest to students and practitioners in physics, mathematics, and engineering, particularly those dealing with geometric calculations and applications in optics and astrophysics.

philipc
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I'm looking for a long list of trig equations.
Mainly as of right now I'm looking for the equation to find the surface area of a sector of a sphere?
Any help would be great,
Thanks
Philip
 
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s = r*θ ; where θ is the angle of sector s.
 
IIRC area of a sphere is 4πr2.

One way to calculate the area of a sector is to determine the fraction of the surface of the sphere that it is.

There are alternatives, but they either involve shape-specific forumlas, or calculus.
 
Ever tried measuring angles on a sphere? You need solid angles for spherical surfaces. The formula you gave is for circles.

The area of the sector depends on how you segment the surface. A few are given here

http://mathforum.org/dr.math/faq/formulas/faq.sphere.html

The one I think you were after is 'sector of a sphere' which is just a simple extension of the circular case.
 
I need to calculate how much light will hit an aperture from the stars above. The area from the aperture will in a since form a region on the outer sphere.
Calculus I can handle, but not sure how to set up the problem.
Thanks
Philip
 
Area of a sphere:

(4 * pi * r^2) / θ

where θ is in radians.
 
Thanks everyone for your help.

Lonewolf,
Due to the fact I nearly made my second post at the exact time you made yours, I didn't see it. But the link you gave me was great thanks. You got me going on the right track with the solid angle, I later found to be called a "steradian", sorry it's been over 10 years since I've had high school Geometry, so I barely know left from right :)
So I've found Area to be Omega = A/r^2, and this helps solve my physics problem.
Again thanks everyone for your help.
Philip
 
I hope you really found the answer,... or are you fooling yourself?

[tex]\omega[/tex] usually represents angular velocity in physics.

If you meant to look for angular displacement, which can be considered area of a sector of a circle if u integrated

[tex]\theta = l/r[/tex]
[tex]l = \theta * r[/tex]
 
Solid angles are often represented by Ω.

Anyways, don't feel bad you didn't know much about this; I don't think I've encountered the term "solid angle" or "steradians" in any of my math courses; they're things I've only picked up on the side.
 
  • #10
I came across solid angles in stellar astrophysics, incidentally, and later in EM. Wonder if they're of any use to mathematicians?
 
  • #11
Good, I don't feel so stupid now [?] thought maybe I was just sleeping in class that day or something :)

So I might see this again in EM? I've heard it's a tuff class, how does it compare to Physics II, because it's taking at least 20+ hours a week for me to get through this stuff, and the stories I hear about EM, well I'm sure you've heard them too.
Philip
 
Last edited:
  • #12
I'm not sure how the courses compare across the atlantic. We don't 'major' in anything like you guys. We pick the degree we're going to take sometime in high school. When we get to university, there are some modules that are compulsory, while there are options related to the chosen degree.
20+ hours a week for a physics course sounds about right though. Depends what's involved in Physics 2. EM was one of the toughest last year. The good news is, once you get through it, things seem to become easier.
 
  • #13
More Trig questions,
how does arccot(x) and slope of a ray hitting a parabolic surface go together?


Lonewolf,
My Physics II class is mostly Electricity and Magnetism, but now we are hitting up optics.
It's been very interesting, but a bunch of new stuff for me.

Philip
 
  • #14
Derivative of [tex]Cot^-1(x)[/tex] looks like what you described.
 
  • #15
Try plotting a parabola against [tex]Cot^-1(x)[/tex] and compare the two graphs. Then think about how rays reflect on parabolic mirrors and such.

A lot of the new stuff you learn will be repeated throughout the degree. They're most likely to be 'breaking you in' during your first few courses. Just keep at it, and you'll be rewarded for it.
 

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