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Hole in a sphere

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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 [the center of] a sphere.

What is the volume of the remaining portion of the sphere?

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

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

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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?

post-5366-1205679362_thumbjpg

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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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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
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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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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

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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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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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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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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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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.

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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Then it wouldn't be a sphere anymore because of the extra sides that were added.

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Then it wouldn't be a sphere anymore because of the extra sides that were added.

That is correct.

Now, find the volume. B))

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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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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

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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?

You have the right idea, exactly.

The hole's length is the length of the remaining volume.

A large portion [as in the case of the earth] of the sphere may have been removed.

No, that formula is not right.

For one thing, volume would have to involve a length cubed, not squared.

To get the correct formula, look the calculations in some of the recent posts,

or just assume the original sphere had a radius of 3" and nothing was removed.

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i'm probably wrong, but i'm 13, and here's what i'm thinking...

i'm assuming that when you said through the sphere, you meant all the way through.

if this is true wouldn't the volume be (6xpi-diameter of hole) which would be 6*3.14=18.84 minus the diameter of the hole.

i hole i'm not completely wrong

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i'm probably wrong, but i'm 13, and here's what i'm thinking...

i'm assuming that when you said through the sphere, you meant all the way through.

if this is true wouldn't the volume be (6xpi-diameter of hole) which would be 6*3.14=18.84 minus the diameter of the hole.

i hole i'm not completely wrong

Hi Billybob,

You're thinking along the right lines, but it's either [1] a little more complicated, or [2] it's a little simpler.

Here's what I mean by that.

[1] The 6 in your equation probably means to you the length of the hole.

If so, your formula has the units of length.

To express volume, you need an expression with units of length cubed: length x length x length.

Doing it this way, you'd need the volume of the original sphere [4pi/3] R3 where R is the sphere's radius,

and then subtract the volume removed by the drill. That's the complicated part.

If you can visualize it, there's a cylinder of length 6" and two spherical "caps":

The volume of the cylinder is easy; the caps are a little harder.

Find them here.

Then it's just arithmetic.

[2] The easier way is to recognize that the problem did not say how big the sphere was.

So you can infer it doesn't matter how big it is.

So you choose an easy size to work with: a diameter of 6"!

Then if you drilled a hole with the smallest imaginable diameter, you'd have a 6" hole that removed no volume at all!

So the remaining volume would simply be the original volume = [4pi/3] 33 = 36pi cubic inches.

If a picture would help, look at this post or simply http://brainden.com/forum/uploads/monthly_02_2008/post-3375-1203670267.png.

Hope that helps. ;)

This puzzle has challenged a lot of smart people.

- bn

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Hope that helps. ;)

This puzzle has challenged a lot of smart people.

- bn

It's in the wording, I really think you need a picture for this...I just spent 4 hours trying to figure out where you guys were getting the 6 inch hole that could magically go from end to end of the earth. You don't DRILL 6 inches is the key part here...The remaining portion that forms a hollow cylinder is 6 inches in length. Which, of course the radius is irrelevent if you always have a condition like that!

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It's in the wording, I really think you need a picture for this...I just spent 4 hours trying to figure out where you guys were getting the 6 inch hole that could magically go from end to end of the earth. You don't DRILL 6 inches is the key part here...The remaining portion that forms a hollow cylinder is 6 inches in length. Which, of course the radius is irrelevent if you always have a condition like that!

Think of a napkin ring - of some diameter - that's 6" high.

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mabey 3*3.14=9.42 9.42 diamater of hole?

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This puzzle was in Scientific American 25 years ago. It requires some lateral thinking. Since the diameter of the drill hole isn't given then it may be assumed that it is not needed for the solution. And one may assume that the puzzle has a solution, appearing in such a prestigious publication. So if the hole diameter isn't important one can assume that the answer is the same for any hole diameter. Assuming an infinitely small diameter hole the answer is 36 pi. And also the answer for all 6" holes through all spheres.

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This puzzle was in Scientific American 25 years ago. It requires some lateral thinking. Since the diameter of the drill hole isn't given then it may be assumed that it is not needed for the solution. And one may assume that the puzzle has a solution, appearing in such a prestigious publication. So if the hole diameter isn't important one can assume that the answer is the same for any hole diameter. Assuming an infinitely small diameter hole the answer is 36 pi. And also the answer for all 6" holes through all spheres.

Thanks. Good to know that. :huh:

It was also on an exam at MIT.

So it's been around, as it should - it's a good one.

It can be calculated.

But lateral thinking came to my rescue when it was posed to me

sometime around 1982 by an MIT grad and colleage who'd had to

answer it as an undergrad.

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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

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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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