Question about turning a hemisphere into an equivalent circle?

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Discussion Overview

The discussion revolves around the concept of transforming the surface of a hemisphere into an equivalent circle, particularly in the context of finding the volume of a sphere and comparing areas. Participants explore the implications of stretching a hemisphere's surface and the resulting geometric properties.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if a hemisphere is covered with a pliable cover and then stretched flat, the resulting circle's radius could be calculated based on the hemisphere's circumference.
  • Another participant corrects the initial claim, stating that the area of the resulting circle after stretching would actually be larger than the surface area of the hemisphere due to the stretching process.
  • The surface area of the hemisphere is noted to be ##2\pi r^2##, while the area of the circle formed from the stretched cover is calculated to be approximately ##2.47 \pi r^2##.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views on the implications of stretching the hemisphere's surface and the resulting area calculations.

Contextual Notes

There are unresolved assumptions regarding the nature of stretching the surface and its impact on area calculations, as well as the definitions of equivalent areas in this context.

goldust
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I was thinking about this when I was trying to work out a simpler way of finding the volume of a sphere. Suppose we cover a hemisphere with a piece of pliable thin cover. Stretching the cover flat would make a circle. The circumference of the sphere is 2*pi*r. The distance along the surface of the hemisphere from the top to the bottom is a quarter of the circumference of the sphere, or 0.5*pi*r. Is it correct that the radius of the circle after stretching out the cover is also 0.5*pi*r? Many thanks in advance!
 
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My bad, I meant turning *the surface of* a hemisphere into an equivalent circle :redface:
 
goldust said:
I was thinking about this when I was trying to work out a simpler way of finding the volume of a sphere. Suppose we cover a hemisphere with a piece of pliable thin cover. Stretching the cover flat would make a circle. The circumference of the sphere is 2*pi*r. The distance along the surface of the hemisphere from the top to the bottom is a quarter of the circumference of the sphere, or 0.5*pi*r. Is it correct that the radius of the circle after stretching out the cover is also 0.5*pi*r? Many thanks in advance!

goldust said:
My bad, I meant turning *the surface of* a hemisphere into an equivalent circle :redface:
And finding the area of the resulting circle?

This won't work. The surface area of the hemisphere is ##2\pi r^2##. If you take the thin cover off the hemisphere and stretch it out, it's true that you get a circle of radius ##(1/2)\pi r##. The area of that circle is ##\pi (\frac{\pi r}{2})^2 = \frac{\pi ^2}{4} \pi r^2 \approx. 2.47 \pi r^2##. I wrote the area in this form to make it easier to compare with the surface area of the hemisphere.

The area of the flattened hemisphere skin is larger than the surface area of the hemisphere, because in flattening the skin, you would need to stretch it, which adds area.
 
Very interesting. Many thanks! :biggrin:
 

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