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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

### #81 bonanova

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Posted 17 March 2008 - 09:11 PM

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!!

If you want to save some calculation time, note that
V[removed] = V[cylinder] + 2xV[spherical cap]

See what you get ...
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### #82 Shae'tan

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Posted 28 March 2008 - 09:16 AM

the quite simplified calculus

pi*integral from -3 to 3 of (-x^2 + 9)=pi*(-x^3/3 + 9*x) evaluated from -3 to 3=pi*(-3^2 + 27 - (3^2 - 27))=pi*(54 - 18) = 36*pi

the long way

sphere
pi*( int{-R,R} [ ((R^2 - x^2)^(1/2))^2 ] )

caps
pi*( int{-R,-3} [ ((R^2 - x^2)^(1/2))^2] ) + pi*( int{3,R} [ ((R^2 - x^2)^(1/2))^2] )

cylinder
pi*( int{-3,3} [ (R^2 - 9)^(1/2))^2] )

integrate

sphere
pi*(x*R^2 - (x^3)/3 |{-R,R}) = pi*(R^3 - (R^3)/3 - (-R^3 - ((-R)^3)/3))

caps
pi*(x*R^2 - (x^3)/3 |{-R,-3}) + pi*(x*R^2 - (x^3)/3 |{3,R}) = pi*(-3*R^2 - ((-3)^3)/3 - (-R^3 - ((-R)^3)/3)) + pi*(R^3 - (R^3)/3 - (3*R^2 - (3^3)/3))

cylinder
pi*(x*R^2 - x*9 |{-3,3})=pi*(3*R^2 - 3*9 - (-3*R^2 - (-3*9)))

algebra

sphere
pi*(2*R^3 - 2*(R^3)/3) = pi*(4*(R^3)/3)

caps
pi*(-3*R^2 + 3^2 + R^3 - (R^3)/3) + pi*(R^3 - (R^3)/3 - 3*R^2 + 3^2) = pi*(4*(R^3)/3 - 6*R^2 + 2*3^2)

cylinder
pi*(6*R^2 - 54)

volume of remnants of sphere = sphere-caps-cylinder
V=pi*(4*(R^3)/3 - 4*(R^3)/3 + 6*R^2 - 2*(3^2) - 6*R^2 + 54) = pi*(54 - 18) = 36*pi

note:
int{a,b} [f(x)] means the integral of f(x) evaluted from a to b
|{a,b} means evaluated from a to b[/font]
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### #83 jason81

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Posted 28 March 2008 - 04:23 PM

I completely understand the question being proposed and I think the bulk of the misunderstanding comes down to semantics.
Some readers are thinking that "drilling a 6 inch hole" means: "Start drilling and only move the drill 6 inches deep."
However, the way the question reads, it actually means: "Drill until the resulting hole is only 6 inches deep."

Understanding this difference makes all the difference.

In the OP, the drill did not move only 6". In fact, the drill actually moved the entire radius of the sphere, through the center. Knowing that the diameter of the drill bit is entirely dependent upon the radius of the sphere. It doesn't matter what the sphere's radius is, or what the diameter of the drill bit is, the "hole" will always be 6" deep, although the drill bit could have moved much further.
This is how the answer can be formulated from the volume of a sphere that has a 6" diameter; the drill bit would have a diameter of 0. As the radius of the sphere approaches the length of the resulting hole, the diameter of the drill bit approaches 0.

As a computer programmer, logic / problem solving is my profession and I love puzzles/problems.
I love this site, it's full of fun puzzles! :-)

--Jason
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### #84 Shae'tan

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Posted 29 March 2008 - 08:07 AM

that calculus proved that it was true for all radii and also gives us a formula you could use for any distance which is pi(L^3)/6. L = the entire length of the hole, in the question this was 6. One could use simple algebra to manipulate this to deal with other measurements. For the reduced L, which is 3, the new formula is pi*8*(L^3)/6.
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### #85 roter

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Posted 09 April 2008 - 06:37 PM

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.

Maybe I am crazy but if you do that then the sphere is gone so the answer is zero, nothing, nada, zilch
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### #86 bonanova

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Posted 09 April 2008 - 07:04 PM

Maybe I am crazy but if you do that then the sphere is gone so the answer is zero, nothing, nada, zilch

Why would the sphere be gone?
What if it were a shorter hole and the sphere ended up looking like a pearl on a necklace?
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### #87 patoruzu

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Posted 09 April 2008 - 10:36 PM

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?

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.

Wordblind: you have all my respect. Very elegant math! Liked the way you used the "donut" to integrate to the entire volume. Congrats!!!
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### #88 giterdone

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Posted 09 April 2008 - 11:33 PM

Spoiler for If a hole was drilled into a sphere...

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### #89 bonanova

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Posted 10 April 2008 - 07:07 AM

Spoiler for If a hole was drilled into a sphere...

That is correct.
Now, find the volume.
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### #90 Budouka

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Posted 22 April 2008 - 01:29 AM

The cylinder is what is removed from the sphere.
Think of beads on a string.
The string is the cylinder.

I finally think I understand where my confusion is coming from on this. The total volume removed from the sphere=the volume of the cylinder+the volume of the endcaps, yes? And so, were one to drill a "six inch hole" in the earth, the actual CYLINDER would have a length of six inches, and the "drill bit" would still have to be long enough to go all the way through, leaving a six-inch-thick ring with an arc-shaped cross section. Am I getting it? I had been envisioning the result of a six-inch drill bit prior to this revelation. But if what I wrote above is correct (I think it must be), then to say "the cylinder is what is removed from the sphere" isn't completely true, because the end-caps are also removed.
I'm still impressed by your story, though.

If the length of the cylinder is "L," is the remaining volume piL^2?
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