Volume of a Semicircular Area? HELP

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To find the volume of a semicircular area with a radius of 2000 km and a thickness of 3000 m, the formula used is (1/2)πr² multiplied by the thickness. The area of the semicircle is calculated as (1/2)π(2000000)², and the volume is then derived by multiplying this area by the thickness of 3000 m. There is some confusion regarding the radius conversion, with one participant noting a potential error in the radius value. The discussion highlights the importance of careful calculations, especially when dealing with large numbers and unit conversions. Overall, the conversation emphasizes the need for accuracy in mathematical problems related to volume calculations.
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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.
 
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Half of an ice cap? If you want to deal with the curvature of the earth, it gets a bit more complicated.

\frac{\pi 20000000^2}{2} 300000}
 
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.
 
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