Question about turning a hemisphere into an equivalent circle?

In summary, the conversation discusses a method for finding the volume of a sphere by covering a hemisphere with a piece of pliable thin cover and stretching it out to form a circle. The circumference of the sphere is 2*pi*r and the distance along the surface of the hemisphere from top to bottom is 0.5*pi*r. However, it is noted that the area of the flattened hemisphere skin is larger than the surface area of the hemisphere due to stretching.
  • #1
goldust
89
1
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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  • #2
My bad, I meant turning *the surface of* a hemisphere into an equivalent circle :redface:
 
  • #3
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.
 
  • #4
Very interesting. Many thanks! :biggrin:
 
  • #5


Yes, it is correct that the radius of the circle after stretching out the cover would also be 0.5*pi*r. This is because when the hemisphere is flattened into a circle, the distance from the top of the hemisphere to the bottom becomes the radius of the circle. Since the distance along the surface of the hemisphere from the top to the bottom is 0.5*pi*r, this would also be the radius of the equivalent circle. This method can be used to find the volume of a sphere by using the formula V = (4/3)*pi*r^3. By finding the area of the equivalent circle, we can then use the formula for the area of a circle (A = pi*r^2) to find the volume of the sphere. This simplifies the calculation and can be a useful approach for finding the volume of a sphere. I hope this helps answer your question.
 

Related to Question about turning a hemisphere into an equivalent circle?

1. How can a hemisphere be turned into an equivalent circle?

The process of turning a hemisphere into an equivalent circle involves finding the radius of the hemisphere, then using that radius to find the circumference of the circle. The circumference of the circle will be equal to the perimeter of the base of the hemisphere.

2. Is it possible to turn any hemisphere into an equivalent circle?

Yes, it is possible to turn any hemisphere into an equivalent circle as long as the radius and height of the hemisphere are known. The formula for finding the circumference of a circle is 2πr, where r is the radius of the circle. By using this formula, we can find the circumference of the equivalent circle for any hemisphere.

3. What is the purpose of turning a hemisphere into an equivalent circle?

Turning a hemisphere into an equivalent circle can be useful for certain calculations and measurements. For example, if you are trying to calculate the surface area or volume of a hemisphere, it may be easier to use the formulas for a circle instead of a hemisphere. It can also help with visualizing and understanding the shape and dimensions of a hemisphere.

4. Can a hemisphere and an equivalent circle have the same volume?

Yes, a hemisphere and an equivalent circle can have the same volume. This is because the volume of a hemisphere is equal to two-thirds of the volume of a cylinder with the same base and height. Since a cylinder and a circle have the same base, they can have the same volume if their heights are equal.

5. Are there any other shapes that can be turned into an equivalent circle?

Yes, there are other shapes that can be turned into an equivalent circle. For example, a cone can be turned into an equivalent circle by finding the radius of the base and using it to calculate the circumference of the circle. Additionally, a spherical cap (a portion of a sphere cut by a plane) can also be turned into an equivalent circle by finding the radius of the base and using it to calculate the circumference of the circle.

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