Volume of a Semicircular Area? HELP

  • Thread starter Thread starter NanoTech
  • Start date Start date
  • Tags Tags
    Area Volume
Click For Summary

Homework Help Overview

The discussion revolves around calculating the volume of a semicircular area, specifically related to the thickness of polar ice caps. The original poster presents a scenario involving a semicircle with a radius of 2000 km and a thickness of 3000 m, seeking the volume in cubic centimeters.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the general formula for volume involving area and thickness, with some exploring the implications of curvature on the calculations. Questions arise regarding the conversion of the radius and its accuracy, as well as notation issues.

Discussion Status

The conversation is active, with participants offering different perspectives on the calculations and questioning the accuracy of the radius conversion. Some guidance on the volume formula has been provided, but no consensus has been reached regarding the specifics of the radius or the overall approach.

Contextual Notes

There are indications of confusion regarding unit conversions and the complexity introduced by the curvature of the Earth. Participants express uncertainty about the notation and the accuracy of their calculations.

NanoTech
Messages
62
Reaction score
0
Hello everyone- My question is about finding the volume of a semicircular area. The radius of the semicircle is 2000km, and the thickness is 3000m? It wants cubic centimeters of ice. (this is about how thick the polar ice caps are). Thanks. ~David W.
 
Physics news on Phys.org
Half of an ice cap? If you want to deal with the curvature of the earth, it gets a bit more complicated.

[tex]\frac{\pi 20000000^2}{2} 300000}[/tex]
 
In general the volume of anything with a constant thickness is
"area times thickness".

For a semi-circle of radius r, the area is (1/2)pi r2.

Putting a thickness of h on that, the volume is (1/2)pi h r2.
 
ok, thanks.
 
Hi all, I have a question about the radius. Is it just me or is the conversion of the radius off by 10. I've done this problem before and I had [(pi*200,000,000)^2/2]300000. Sorry about the notation, I'm new at this.
 
Originally posted by boxelderbug
Hi all, I have a question about the radius. Is it just me or is the conversion of the radius off by 10. I've done this problem before and I had [(pi*200,000,000)^2/2]300000. Sorry about the notation, I'm new at this.

Yeah, I insert small random errors to... er... make sure that people check the work, yeah that's it. Honestly, the problem is that reading problems and lot's of zero's don't mix well.
 

Similar threads

Replies
1
Views
1K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
4
Views
1K
Replies
12
Views
3K
  • · Replies 15 ·
Replies
15
Views
58K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K