A hemisphere rests with its circular base on a horizontal, level surface, and is to be cut into two pieces of equal volume. If the hemisphere’s radius is r, at what fraction of r above the floor should the horizontal cut be made?

[Solution in next post: https://robertlovespi.wordpress.com/2015/04/06/working-towards-a-solution-of-the-hemisphere-problem/]

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i going to give it more thought tomorrow!

Leslie

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(correction) I’m going to give it more thought tomorrow. I think I must be tired.

Leslie

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Let the volume of the hemisphere be 2Pi*r^3

The problem requires this to be truncated into two equal volume shapes

(1) a spherical prism with its one face same as the radius of the original sphere and the other face having the radius R1 where R1 is unknown.

(2) a new hemisphere with base radius of R1 and volume Pi*r^3

For the new hemisphere 4*Pi*R1^3 = Pi*r^3

or 4R1^3 =r^3

or R1 = r/(4)^1/3

If r = 1

then R1 = 0.63287 approximately

Knowing R1, H, the truncation height can be found by Pythagoras theorem.

Ashok

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I’m able to follow some of that, but wouldn’t the original hemisphere’s volume be (2/3)(pi)(r^3), or only a third what you used, since a sphere’s volume is (4/3)(pi)(r^3)?

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You are right that the volume of a sphere is 4/3 Pi *r^3.

The reasoning that follows wont change but the final answer will be still be the same as the error is made on both side of the equation. A corrected solution is pasted below

Let the volume of the hemisphere be 2/3Pi*r^3

The problem requires this to be truncated into two equal volume shapes

(1) a spherical prism with its one face same as the radius of the original sphere and the other face having the radius R1 where R1 is unknown.

(2) a new hemisphere with base radius of R1 and volume 1/3Pi*r^3

For the new hemisphere 4/3*Pi*R1^3 = 1/3Pi*r^3

or 4R1^3 =r^3

or R1 = r/(4)^1/3

If r = 1

then R1 = 0.63287 approximately

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For the height of the horizontal cut (which R1 is not, correct?), does this yield a solution which matches the approximate solution in the next post, or are these two answers contradictory?

Your term “spherical prism” confuses me. How do you find the volume of such a thing, other than by subtracting the volume of a “spherical cap” (explained in the next post) from that of a hemisphere?

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Dear Robert,

The absolutely correct answer is very close to your answer that is (1-0.6238270 =0.37613 away from the original base.

Its very easy to see that the line joining the outer edge of base radius to the vertex, forms a isosceles right angle triangle as the two sides are the radius of the original sphere.

Thus the parting line with a radius of 0.6 etc will also be 0.6etc down from the vertex or 0.37613 up from the base.

I am ignorant as to how to find the volume of a spherical prism.

In the problem posed , it was already given.

Do you want me to find out the formula?

PS

When you have time when do not you write in detail how you have used Stella to make any of the polygon that you make.

I have had Stella for years but never learnt how to make such solids.

Regards

Ashok

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Hm. I’m not sure what is causing the small discrepancy in our answers; perhaps someone else will be able to help resolve it. As for “spherical prisms,” I had not encountered that term before you used it in an earlier comment.

Regarding

Stella, have you tried the on-line “help” /Stellatutorial material? I thinkStella‘s author may have also put some tutorials on YouTube. The main thing withStellais to just sit down and do stuff with it; you’ll learn as you experiment with it. On some posts of polyhedra on this blog, there are step-by-step explanations of how I made those polyhedra, but those posts are in the minority, I’m afraid.LikeLike

Ashok, here is one such post where I did describe just what I did with

Stellato produce the featured polyhedron. https://robertlovespi.wordpress.com/2014/11/18/an-alteration-of-the-icosahedrondodecahedron-compound/ There are other such posts, but finding them . . . well, I should have set that up to be easier, but it didn’t occur to me to do so.LikeLike

The next post shows an approximate solution (~35% of the way from the ground to top): https://robertlovespi.wordpress.com/2015/04/06/working-towards-a-solution-of-the-hemisphere-problem/

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Turn it upside down and you can ask what the water level of a full hemispherical bowl will be when half of it is drunk. đź™‚

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