# Escape the “bowl” above the abyss [closed]

You are at the bottom of this sphere section. You know the two given measures. The “bowl” is above and in the middle of circular opening of an abyss. The opening is circular with a radius of 4.87. The spherical shape may be flattened by you into a circular disk that should cover the abyss opening to allow you escape. Is it possible?

## closed as unclear what you're asking by elias, BmyGuest, Gamow, Beastly Gerbil, IAmInPLSAug 22 '16 at 12:35

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• It is not clear what assumptions can be made about the way the material behaves when it is deformed. Does the total area stay the same? Or does the material stretch transversly but not radially? Once that assumption is filled in, it seems a straightforward maths problem. – Jaap Scherphuis Aug 22 '16 at 5:38
• Not sure I got the question completely: essentially we should use the bowl surface area to cover/extend over the abyss hole? Is coverage needed, or can just two 'stretches' anchor the rest on two sides? – BmyGuest Aug 22 '16 at 5:40
• Is it allowed to 'bend' the material? I.e. Can one make a strip and fold it over at the end to create a longer strip? – BmyGuest Aug 22 '16 at 5:42
• The area of the final disk is equal the area of the spherical section - in other words, the material is such that it may be flattened gradually. Creating strip will not cover the abyss opening - the goal is to cover the abyss opening. So just two anchors solution will not cover the abyss. – Moti Aug 22 '16 at 5:46
• But the spherical section is 3d - it has volume. You can make it arbitrarily thin and it'll always work. – Deusovi Aug 22 '16 at 5:53

I interpret this question as asking:

Is the surface area of the bowl shape greater than the surface area of a disk with radius $4.87$?

To answer this, we must first determine the radius of the sphere:$$r^2=(r-4)^2+3^2\\0=25-8r\\r=\frac{25}8$$

(Note that this is less than 4, so the "bowl" is actually more than half a sphere and starts closing back on itself at the top.)

Now we use the fact that the surface area of a section of height $h$ of a sphere with radius $r$ is $2\pi rh$. Therefore the surface area of the bowl is $2\pi\times\frac{25}8\times4=25\pi$.

This is same as the surface area of a disk with radius $5$, so the bowl is large enough to cover the opening.

• This is a good answer – Moti Aug 22 '16 at 16:15