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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

### #101 bonanova

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Posted 14 June 2008 - 08:23 AM

Does this hold for offset holes?

I'm neither a mathematician, nor much of a logician. But when I thought of this puzzle, I refused to think on so large of a scale. I imagined a 12" sphere with an offset hole. I imagined the shape that would be drilled out, and considered that 6" length, was that the longest length, the shortest, or the average. At that point I could no longer go on.

Sigh.
Tyg

Don't know what you mean by offset, but the hole is drilled though the center of the sphere.
Does that help?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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### #102 PickleLlama

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Posted 23 June 2008 - 07:41 PM

Spoiler for spoiler courtesy of bonanova

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### #103 bonanova

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Posted 24 June 2008 - 07:54 AM

Spoiler for spoiler courtesy of bonanova..
Nice job.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #104 Interested

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Posted 15 July 2008 - 05:51 PM

Apart from the degree of mathematical and logical aptitude, I praise you, bonanova, for your patience!

I agree 100%!!! It took me 46 posts to get it although if jason81 would have added his answer sooner I may have been much faster on the uptake. I am just proud that I got it at all as I was loosing hope.

Thank you again Bonanova for your patience. We need more teachers like you in our school system...

Not to bite off of Jkyle1980, but where do I send my tuition as well...?
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### #105 Aiken38

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Posted 15 July 2008 - 07:30 PM

36pi, nice.. i started working out the mathmatical solution and then it got me thinking and then i figured it out.. cool!
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### #106 AliceJH

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Posted 01 August 2008 - 04:37 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.

Edited by AliceJH, 01 August 2008 - 04:39 PM.

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### #107 bonanova

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Posted 01 August 2008 - 04:42 PM

I agree 100%!!! It took me 46 posts to get it although if jason81 would have added his answer sooner I may have been much faster on the uptake. I am just proud that I got it at all as I was loosing hope.

Thank you again Bonanova for your patience. We need more teachers like you in our school system...

Not to bite off of Jkyle1980, but where do I send my tuition as well...?

Hi interested,

Thanks for the kind words.
Hope you're enjoying Brain Den as much as I do.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #108 xucam

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Posted 11 November 2008 - 08:25 PM

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.

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.

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'.

There was another purported 'solution' which seems to me to be flawed:

You're all making this too hard. You just need to know the area of each slice of the donut, which is simply the difference of two circles.

L = length of bore
x = radius of sphere cross-section at height z
A = area of donut cross-section at height z

Forming two simple right triangles:
r? = R? - L?/4
x? = R? - z?

The first equation stipulates that the radius of the cylinder equals the radius of the sphere minus (the length of the bore divided by 4), which means that
r = R - 1.5 which could be true for one such case, but is hardly a general solution - if the difference between the radii of the cylinder and sphere were 1.5 inches then the volume would be way more than 36 pi cubic inches.

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

But still, great for you that you felt like a genius and thanks for sharing.

Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way.

A 6-inch [long] hole is drilled through a sphere.
What is the volume of the remaining portion of the sphere?

The hard way involves calculus. The easy way uses logic.

... combining 2 posts by the author ...

The easy way is to suppose the answer is the same for
any sphere [with diameter not less than 6 inches], and
calculate the answer for a 6-inch diameter sphere.

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### #109 AfroAllianceRRR

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Posted 11 November 2008 - 11:33 PM

6(pi)rsquared.

Bssically just putting it down as an expresion of the voloume of the drilled hole equals the empty space of the the sphere, I think it might be unorthodox but it worth a shot
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### #110 bonanova

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Posted 01 April 2009 - 05:38 PM

6(pi)rsquared.

Bssically just putting it down as an expresion of the voloume of the drilled hole equals the empty space of the the sphere, I think it might be unorthodox but it worth a shot

Not quite.
There are three volumes to consider.
• The cylindrical hole
• The spherical end caps on either end of the hole cylinder
• The remaining volume of the sphere.
It's the third item we seek.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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