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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


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#21 Wordblind

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Posted 26 August 2007 - 09:50 AM

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
r = radius of bore
R = radius of sphere
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?

Using the formula for the area of a circle
A = ?(x?-r?) = ?(L?/4-z?)
The area of the cross is independent of R, so the full volume is constant.

To find the volume, you can integrate the area from z=-L/2 to L/2, but it is simpler to use the degenerate case.
R=3, r=0
V = 4?R?/3 = 36? cu.in.
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#22 bonanova

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Posted 26 August 2007 - 02:48 PM

Very nice!

There are two steps to get to 36pi.

[1] deducing it doesn't depend on R.
[2] computing the R=3 case.

Step [1] can be done logically or mathematically.

Wordblind absolutely wins the prize for the slickest math.
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#23 cpotting

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Posted 26 August 2007 - 07:21 PM

cpotting,

kudos for finding all these neat [cap and barrel] formulas.
That's does all of the calculus work.

I found it useful to put everything in terms of R and h.
You can do this by noting that RR = rr + hh [Pythagorus]

V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]

V[cylinder] = height x area = Hprr = 2ph [RR - hh]

V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.

Recalling that h=3,

V[barrel] - V[cylinder] = 36p [a constant].

Looking at your derivation, everything is correct.
If you add like terms in your expression for V,
you'll see that all the rr terms add up to zero.

Your derivative expression is correct, also,
except that you should include the constants (1/12)pH8, etc...
[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]
You'll see again that the r-dependent terms add up to zero,
and the derivative is zero.

OK?



Great. Thanks bonanova - I see where I went wrong now.
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#24 roolstar

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Posted 29 January 2008 - 02:52 PM

So you're saying that, if you are a Doughnut baker, making the hole bigger while maintaining the same height will not save you any money...

But still your customers will intuitively feel you're cheap so they'll switch to another bakery....

And the only customers you may keep are the guys on this forum.....

Lovin' it! :D

Edited by roolstar, 29 January 2008 - 02:57 PM.

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#25 McSquid

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Posted 01 February 2008 - 08:00 PM

Im sorry, I still fail to grasp this logical concept. if you drill a hole with a diameter of 0, then have you really drilled a hole? if so you must have done it with an invisible drill bit. Then again i could be missing the point entirely
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#26 bonanova

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Posted 01 February 2008 - 11:35 PM

Im sorry, I still fail to grasp this logical concept. if you drill a hole with a diameter of 0, then have you really drilled a hole? if so you must have done it with an invisible drill bit. Then again i could be missing the point entirely

Spoiler for Here's what is going on .. "the point

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#27 Jkyle1980

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Posted 02 February 2008 - 03:03 AM

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 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.


I am still not seeing this. How can the volume of the two spheres below be the same?

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#28 bonanova

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Posted 02 February 2008 - 08:46 AM

I am still not seeing this. How can the volume of the two spheres below be the same?

Answer this question and you'll understand the puzzle:
How can equal-length cylinders go [completely] through unequal-diameter spheres?

Hint: It's possible to drill a 6-inch-long hole through the earth.
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#29 Jkyle1980

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Posted 03 February 2008 - 08:28 AM

Answer this question and you'll understand the puzzle:
How can equal-length cylinders go [completely] through unequal-diameter spheres?

Hint: It's possible to drill a 6-inch-long hole through the earth.


By not going through the center of the sphere?
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#30 bonanova

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Posted 03 February 2008 - 10:31 AM

By using a drill with a larger diameter.
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