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# Hole in a sphere

## Question

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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First calculate the radius of the hole in terms of the radius of the sphere, R. At the end of the sphere, the vertical leg is 3 inches, the hypotenuse is R, so the radius of the hole is Rh=sqrt(R2-32). Next, calculate the outer circle radius in terms of the height x above or below the center of the sphere and the radius of the sphere. This is Ro=sqrt(R2-x2). The cross-sectional area is then pi Rh2 - pi Ro2 or

pi (R2-x2) - pi (R2-32) =

pi (32-x2) =

pi (9-x2).

the volume is the integral of this area from x = -3 to 3, or

pi (9x - x3/3) {at x = 3} - pi (9x - x3/3) {at x = -3}, or

pi (27 - 9) - pi (-27 + 9) =

pi (27 - 9 + 27 - 9) =

36 pi.

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First calculate the radius of the hole in terms of the radius of the sphere, R. At the end of the sphere, the vertical leg is 3 inches, the hypotenuse is R, so the radius of the hole is Rh=sqrt(R2-32). Next, calculate the outer circle radius in terms of the height x above or below the center of the sphere and the radius of the sphere. This is Ro=sqrt(R2-x2). The cross-sectional area is then pi Rh2 - pi Ro2 or

pi (R2-x2) - pi (R2-32) =

pi (32-x2) =

pi (9-x2).

the volume is the integral of this area from x = -3 to 3, or

pi (9x - x3/3) {at x = 3} - pi (9x - x3/3) {at x = -3}, or

pi (27 - 9) - pi (-27 + 9) =

pi (27 - 9 + 27 - 9) =

36 pi.

Nice job.
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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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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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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
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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 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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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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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.

1. The cylindrical hole
2. The spherical end caps on either end of the hole cylinder
3. The remaining volume of the sphere.
It's the third item we seek.
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You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.

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You guys are idiots. There is no answer. You don't know the information of the sphere to figure this out. The Circumference could be 100 inches or 50 inches you don't know.

Hello coolfflame, and welcome to the Den.

There is enough information to solve this.

You will earn respect from members when you respect them.

That would mean first assuming a puzzle does have an answer and trying to find it.

If in the process you come to believe that it can't be solved, show that in the form of a proof.

Name calling, even in jest, won't earn you points with the site administrator.

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Hi,

I've waded through 12 pages of comments about this puzzle, and I am happy that it is an interesting geometrical puzzle, and that the calculus-based solutions demonstrate that the volume of the remaining 6 inch high donut/bracelet/sliver/whatever shape has a fixed volume of 36pi for all spheres with a diameter greater than 6inches.

However, way back at the start there was the promise of 'logical' solution than did not need calculus. I've not seen it in any of the 12 pages.

Forgive me if I've missed a subtelty somewhere, but the 'logical' solution being offered appears to be based on the premise that the original question does not specify a diameter, therefore the solution does not depend on the diameter, therefore we just have to establish the remaining post-drilled volume for any one particular diameter of sphere, oh look for diameter=6 the answer is 36pi, and voila we have the answer for all diameters.

That's not a complete solution. That's a solution based on a very big, unproven, assumption.

What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place that the solution does not depend on the diameter?

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What is the 'logical', non-calculus-based, reasoning that leads us to be able to state in the first place

that the solution does not depend on the diameter?

Modus ponens

If the puzzle is well posed, then it contains all necessary information.

Following the first rule of puzzle solving, we take the puzzle to be well posed.

Therefore we take that the puzzle contains all necessary information.

If the diameter is necessary information, then it is contained in the puzzle.

It is not contained in the puzzle.

Therefore the diameter is not necessary information.

If the solution depends on the diameter, then the diameter is necessary information.

The diameter is not necessary information.

Therefore, the solution does not depend on the diameter.

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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.

A hole does not have a curved top or bottom.

A curved portion [spherical caps] on either end of the hole is removed when the hole is made.

But inspecting the inner surface of the hole we find a circular cylinder with a well defined length. That length is given as 6".

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xucam raises several issues. Here are the responses, belated.

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.

Quite right. That has been fixed in the OP.

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.

Having thus obtained the solution, there's no need to drill the hole.

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'.

Since the volume of the end caps was not asked, it does not belong in the answer.

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

The is in fact 36pi cu in.

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So. Um. How would you find the volume of a six-inch drill hole? Because you could just use the Volume of the Sphere-Volume of the drill hole to get the answer... Unless of course the idea of if the sphere is too small is added.

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Okay, so I don't think any calculus is required here. It's just simple geometry. Feel free to let me know if I made any mistakes in my calculations:

L = height of cylinder (6)

h = height of cap

Vsphere = 4/3 pi R3

Vcylinder = L pi r2 = 6 pi r2

Vcap = 1/3 pi h2 (3 R â€“ h)

r2 + L2/4 = R2 (based on triangle from center of sphere to edge of cylinder up to top of cylinder)

r2 = R2 â€“ 9 (substituting 6 for L as given in the problem)

h = R â€“ L/2 = R â€“ 3 (this one should be self explanatory)

Therefore, putting everything into terms of R:

Vcylinder = 6 pi (R2 â€“ 9) = 6 pi R2 â€“ 54 pi

Vcap = 1/3 pi h2 (3 R â€“ h) = 1/3 pi (R â€“ 3)2 (3R â€“ (R â€“ 3)) = 1/3 pi (R2 â€“ 6R + 9) (2R + 3) = 1/3 pi (2R3 â€“ 9 R2 +27) = 2/3 pi R3 â€“ 3 pi R2 + 9 pi

Vcylinder + 2 Vcap = 6 pi R2 â€“ 54 pi + 4/3 pi R3 â€“ 6 pi R2 + 18 pi = 4/3 pi R3 â€“ 36 pi

Remaining Volume = Vsphere â€“ (Vcylinder + 2 Vcap) = 4/3 pi R3 â€“ (4/3 pi R3 â€“ 36 pi) = 36 pi

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i was confused for a bit on what the value 6 was corresponding to; once i visualized it, the question was cake

simple solution: just imagine the diameter of the sphere is 6, then the cylinder would have 0 radius and the remaining volume is the volume of the sphere: 4/3pi*R^3 = 36pi

logically, any larger sphere would have the same remaining volume

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The geometric problem has been stated in the form of a poem

(which I must attribute to Anonymous as I could not find the author):

Old Boniface he took his cheer,

Then he bored a hole through a solid sphere,

Clear through the center, straight and strong,

And the hole was just six inches long.

Now tell me, when the end was gained,

What volume in the sphere remained?

Sounds like I haven't told enough,

But I have, and the answer isn't tough!

The answer to the problem is 36*pi cubic inches, the volume of a 6 inch sphere.

Edited by Dej Mar
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I haven't read the whole thread, so I don't know if is this has been posted before...

My thought is that you could submerge the sphere with a hole in it into a graduated glass container with water and just read how much water it displaces. I'm not sure that the wording is correct, but you get what I mean.

edit: grammar

Edited by andromeda
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The geometric problem has been stated in the form of a poem

(which I must attribute to Anonymous as I could not find the author):

Old Boniface he took his cheer,

Then he bored a hole through a solid sphere,

Clear through the center, straight and strong,

And the hole was just six inches long.

Now tell me, when the end was gained,

What volume in the sphere remained?

Sounds like I haven't told enough,

But I have, and the answer isn't tough!

The answer to the problem is 36*pi cubic inches, the volume of a 6 inch sphere.

Dej Mar

Thanks for adding this great poetic version of the puzzle!

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well if we consider the radius of the cylindrical hole to Zero, the answer become obvious..nothing has been cut and the volume of the sphere is (4/3)*pi*3*3*3=36 pi

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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 a sphere.

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

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

What is the circumference of the six inch hole? Wide drill bit, thin drill bit, what?

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I haven't read the whole thread, so I don't know if is this has been posted before...

My thought is that you could submerge the sphere with a hole in it into a graduated glass container with water and just read how much water it displaces. I'm not sure that the wording is correct, but you get what I mean.

edit: grammar

EUREKA! If you know your scientific history, you'll know what I mean. Only problem I see with this is that what is asked for is the answer, not the process to the answer.

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What is the circumference of the six inch hole? Wide drill bit, thin drill bit, what?

for logical way, consider the radius of the hole to be Zero and you get the answer

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