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# Hole in a sphere

Best Answer bonanova, 23 August 2007 - 07:11 AM

The volume of the spherical caps is given by:
[list]
where
[list]
[*] h = the height of the cap (difference between r and the distance from the centre of the[/*:m:1cc31][list] sphere to the centre of the circular end of the hole)

Kudos to cpotting for the cap formula.
Spoiler for Here's the mathematical solution
Spoiler for Here's the logical solution
Go to the full post

166 replies to this topic

### #111 coolfflame

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Posted 30 July 2009 - 06:27 AM

You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.
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### #112 bonanova

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Posted 30 July 2009 - 07:02 AM

You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.

Hello coolfflame, and welcome to the Den.

There is enough information to solve this.

You will earn respect from members when you respect them.
That would mean first assuming a puzzle does have an answer and trying to find it.
If in the process you come to believe that it can't be solved, show that in the form of a proof.

Name calling, even in jest, won't earn you points with the site administrator.

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #113 upthebuzzard

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Posted 15 August 2009 - 09:31 PM

Hi,

I've waded through 12 pages of comments about this puzzle, and I am happy that it is an interesting geometrical puzzle, and that the calculus-based solutions demonstrate that the volume of the remaining 6 inch high donut/bracelet/sliver/whatever shape has a fixed volume of 36pi for all spheres with a diameter greater than 6inches.

However, way back at the start there was the promise of 'logical' solution than did not need calculus. I've not seen it in any of the 12 pages.

Forgive me if I've missed a subtelty somewhere, but the 'logical' solution being offered appears to be based on the premise that the original question does not specify a diameter, therefore the solution does not depend on the diameter, therefore we just have to establish the remaining post-drilled volume for any one particular diameter of sphere, oh look for diameter=6 the answer is 36pi, and voila we have the answer for all diameters.

That's not a complete solution. That's a solution based on a very big, unproven, assumption.

What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place that the solution does not depend on the diameter?
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### #114 bonanova

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Posted 16 August 2009 - 08:51 AM

What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place
that the solution does not depend on the diameter?

Spoiler for Here you go

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #115 bonanova

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Posted 17 August 2009 - 12:22 PM

The question does not state the diameter or length of the hole (a cyclinder with a curved top and bottom) in relation to diameter of the sphere.
This information is essential.
We only know that the diameter of the sphere must be greater than 6 inches.
You can use the same size apple corer for different size apples.

A hole does not have a curved top or bottom.
A curved portion [spherical caps] on either end of the hole is removed when the hole is made.
But inspecting the inner surface of the hole we find
a circular cylinder with a well defined length. That length is given as 6".

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #116 bonanova

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Posted 17 August 2009 - 12:40 PM

xucam raises several issues. Here are the responses, belated.

The problem as originally stated is not solvable, as it could be reasonably interpreted as a 6-inch diameter hole all the way through an object.

Quite right. That has been fixed in the OP.

I would argue that you cannot drill a 6 inch hole through a sphere with radius = 3 any more than you can through a sphere of r = 2. A hole with no volume is not a hole. It may be true that the r=3 case is a boundry which a solution approaches, and obviously the volume of such an intact sphere is 4/3 * 3 * 3 * 3 * pi cu. in.= 36 pi cubic inches.

Having thus obtained the solution, there's no need to drill the hole.

There was also discussion of the 'caps' that would fit on top of the inscribed cylinder in the problem. Clearly, for the example of boring through the earth with a bit whose radius is almost as big as the radius of the earth, these caps would be massive (basically, you've just cut away 6-inch slice of the earth through the center). The volume of these caps is very obviously dependent on the radius of the sphere, and are thus not included in the 'answer'.

Since the volume of the end caps was not asked, it does not belong in the answer.

the stated 'solution' of 36pi isn't a volume at all, whereas 36 pi cubic inches is...

The stated answer is in fact 36pi cu in.

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #117 jsai3334

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Posted 21 October 2009 - 12:59 AM

So. Um. How would you find the volume of a six-inch drill hole? Because you could just use the Volume of the Sphere-Volume of the drill hole to get the answer... Unless of course the idea of if the sphere is too small is added.
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### #118 Scott Strosahl

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Posted 16 April 2010 - 11:57 PM

Okay, so I don't think any calculus is required here. It's just simple geometry. Feel free to let me know if I made any mistakes in my calculations:

L = height of cylinder (6)
h = height of cap

Vsphere = 4/3 pi R3
Vcylinder = L pi r2 = 6 pi r2
Vcap = 1/3 pi h2 (3 R – h)

r2 + L2/4 = R2 (based on triangle from center of sphere to edge of cylinder up to top of cylinder)
r2 = R2 – 9 (substituting 6 for L as given in the problem)

h = R – L/2 = R – 3 (this one should be self explanatory)

Therefore, putting everything into terms of R:
Vcylinder = 6 pi (R2 – 9) = 6 pi R2 – 54 pi
Vcap = 1/3 pi h2 (3 R – h) = 1/3 pi (R – 3)2 (3R – (R – 3)) = 1/3 pi (R2 – 6R + 9) (2R + 3) = 1/3 pi (2R3 – 9 R2 +27) = 2/3 pi R3 – 3 pi R2 + 9 pi
Vcylinder + 2 Vcap = 6 pi R2 – 54 pi + 4/3 pi R3 – 6 pi R2 + 18 pi = 4/3 pi R3 – 36 pi

Remaining Volume = Vsphere – (Vcylinder + 2 Vcap) = 4/3 pi R3 – (4/3 pi R3 – 36 pi) = 36 pi
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### #119 yakul85

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Posted 17 July 2010 - 10:29 AM

i was confused for a bit on what the value 6 was corresponding to; once i visualized it, the question was cake

simple solution: just imagine the diameter of the sphere is 6, then the cylinder would have 0 radius and the remaining volume is the volume of the sphere: 4/3pi*R^3 = 36pi

logically, any larger sphere would have the same remaining volume
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### #120 Dej Mar

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Posted 17 July 2010 - 11:04 AM

Spoiler for I like this puzzle!

Edited by Dej Mar, 17 July 2010 - 11:08 AM.

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