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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

### #71 Knight

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Posted 03 March 2008 - 09:11 AM

What I think is that drilling a hole doesnot reduce the Volume of Sphere. Yes it will increase the density as Surface area will be increased due to drilling of hole., but the volume will remain same.Am I right?

Edited by Knight, 03 March 2008 - 09:12 AM.

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### #72 bonanova

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Posted 03 March 2008 - 10:23 AM

What I think is that drilling a hole does not reduce the [1] Volume of Sphere.
Yes it will increase the [2] density as [3] Surface area will be increased due to drilling of hole,
but the volume will remain same.
Am I right?

[1] The volume of a sphere can be reduced only by reducing its radius, as it remains a sphere.
In the puzzle, the drilled sphere is no longer a sphere - it's like a bead that can be put on a string.
The drill removes some volume of material from the sphere.
The question asks what is the volume of the remaining portion of the sphere.

[2] Mass removed is proportional to the volume removed, so density remains the same.
Density has nothing to do with the puzzle, which is purely geometric.

[3] Surface area does not necessarily increase.
It may actually decrease if the sphere is originally large enough.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #73 Lost in space

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Posted 03 March 2008 - 11:21 AM

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

1 - Are we to assume the axis is the point of entry and exit?
If not size of sphere is not obtainable

2 - diameter of hole is not included, can not deduce the REMAINING volume.

Spoiler for Eureka!!!!!

Spoiler for Size is important!

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### #74 bonanova

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Posted 03 March 2008 - 01:36 PM

1 - Are we to assume the axis is the point of entry and exit?
If not size of sphere is not obtainable

2 - diameter of hole is not included, can not deduce the REMAINING volume.

[1] Axis of drill passes through center of sphere. And out the other side, of course
[2] Yes you can.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #75 Lost in space

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Posted 03 March 2008 - 02:24 PM

Yes you can.

yes if I want to give a formula and not give a figure, I will have to go for displacement/eureka. Will you accept that as a "method/answer"
Not gonna give the lady up without a fight are you
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### #76 bonanova

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Posted 03 March 2008 - 09:58 PM

yes if I want to give a formula and not give a figure, I will have to go for displacement/eureka.
[1] Will you accept that as a "method/answer"
[2] Not gonna give the lady up without a fight are you

[1] Eureka is always good; displacement seems irrelevant. The method is up to you; the OP just asks for the answer.
[2] If you saw the lady, you might become interested. It's entirely up to her; have at it, and good luck.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #77 TheKidUpstairs

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Posted 08 March 2008 - 04:44 AM

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

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Posted 16 March 2008 - 03:58 PM

To get an exact answer we MUST know the diameter of the drilled hole. Without this information, we cannot assume the sphere to be 6 in. in diameter, we cannot know the size of the sphere, we cannot know the volume of the sphere or the cylinder. Look at the attached picture. Am I right....or am I missing something?

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### #79 glabella_fuzz

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Posted 16 March 2008 - 05:00 PM

To get an exact answer we MUST know the diameter of the drilled hole. Without this information, we cannot assume the sphere to be 6 in. in diameter, we cannot know the size of the sphere, we cannot know the volume of the sphere or the cylinder. Look at the attached picture. Am I right....or am I missing something?

Okay, leave it to me, the new guy, to make a post and question my very own reply.

I am still pondering about this and realize that a bigger sphere would mean a bigger hole...........

a bigger hole.......less volume left..............[thinking as i type]..............

Is this really constant?? As the hole gets bigger, the radius of the drilled hole grows........I have yet to manually figure this out with calculus.

Could it really be that no matter how big you make the sphere, the larger hole will always displace enough of the sphere to leave 36pi?

Sacred Bovine!!.....I think i finally get it!!
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### #80 kdawghomie

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Posted 17 March 2008 - 12:00 AM

Okay, leave it to me, the new guy, to make a post and question my very own reply.

I am still pondering about this and realize that a bigger sphere would mean a bigger hole...........

a bigger hole.......less volume left..............[thinking as i type]..............

Is this really constant?? As the hole gets bigger, the radius of the drilled hole grows........I have yet to manually figure this out with calculus.

Could it really be that no matter how big you make the sphere, the larger hole will always displace enough of the sphere to leave 36pi?

Sacred Bovine!!.....I think i finally get it!!

LOL @ "Sacred Bovine"... I thought you typed "Sacred Bonanova" at first XD

But yes I do believe it is logical to think that the increasingly larger hole in the sphere will always displace enough of the sphere to get 36pi.

Edited by kdawghomie, 17 March 2008 - 12:02 AM.

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