Radius of circle obtained by flattening a hemisphere

[goldust]

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]

My bad, I meant turning *the surface of* a hemisphere into an equivalent circle

 

[Mark44]

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!

My bad, I meant turning *the surface of* a hemisphere into an equivalent circle

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.

 

[goldust]

Very interesting. Many thanks!