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

You can't really assume anything. And a hole is defined as an "unoccupied space." Hardly fits.

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

The circumference (and thanks for teacing me that word) depends on the size of the sphere. It will be different for different sizes of spheres. And despite your reservations, it was a good advise you got, to assume a zero circumference. Even if it, technically, won't be a real hole.

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No, because for instance the drill bit used to drill the hole could vary dramatically, thus affecting the volume taken away from the sphere.

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For each size of sphere, there is only one circumference that fits the conditions of the riddle. Imagine a sphere the size of the earth. The circumference will have to be huge, if the hole needs to be 6 inches long.

My point is, while the volume taken away from the sphere depends on the size of the sphere, the volume of what's left is constant.

No, because for instance the drill bit used to drill the hole could vary dramatically, thus affecting the volume taken away from the sphere.

I once read a similar two-dimentional riddle, I'll try to find it and post it.

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No no no, circumference wouldn't vary with size of original object. Volume, however, definitely would. Circumference is the distance around a circle. And your spoiler just doesn't make sense to me.

Edited by NickFleming
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Okay, we're missing two factors we need; The original volume, and the diameter/circumference of the hole. So I'll just write it in terms of the two.

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First, find volume of original sphere. volume of sphere=2/3 volume of smallest cylinder it could have been cut from. Assuming 6 inches is all the way through the hole, the diameter of sphere, and therefore the original cylinder, is 6 inches. so to find volume of cylinder, we set up pi times 32 times 6. so then pi times 54. I'll just leave it in terms of pi to be most exact. So, Archimedes found out that volume of sphere=2/3 that of cylinder. So now we have 32pi=volume of sphere. So answer would be 32p-(6pr^2), p being pi, and r representing missing radius of hole. This is because missing volume would be much different depending on whether diameter of hole was 6m or 600.

Restatement of answer: 32p - 6pr2

Any problems with my logic, tell me. And yes, I kno I found volume of sphere different than most people would. Turned out we didn't need original volume.

Edited by NickFleming
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It wd seem that knowing the size of the drill bit wd be helpful.

(the volume of the sphere remaining after drilling a hole with a 1/32" drilling or a 1/2" drill bit wd be quite disparate)

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I think I understand what they meant, but I could be wrong.

In the picture, the holes are both 6 inches long(the dark gray part), but there are caps (the light gray part) that are cut off.

So only the white part is left and that's the part we are looking for.

That's why they don't tell you the diameter. It changes depending on the size of the sphere.

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First, find volume of original sphere. volume of sphere=2/3 volume of smallest cylinder it could have been cut from. Assuming 6 inches is all the way through the hole, the diameter of sphere, and therefore the original cylinder, is 6 inches. so to find volume of cylinder, we set up pi times 32 times 6. so then pi times 54. I'll just leave it in terms of pi to be most exact. So, Archimedes found out that volume of sphere=2/3 that of cylinder. So now we have 32pi=volume of sphere. So answer would be 32p-(6pr^2), p being pi, and r representing missing radius of hole. This is because missing volume would be much different depending on whether diameter of hole was 6m or 600.

Restatement of answer: 32p - 6pr2

Any problems with my logic, tell me. And yes, I kno I found volume of sphere different than most people would. Turned out we didn't need original volume.

I didn't understand all of your answer, but I have a small remark. 6 inches is the hight of the hole's "wall", not neccesariy the diameter of the sphere. Actually, if the size of the hole was zero, the sphere would be 6 inches diameter.

Hope I uploaded my sketch correctly. I know t8t8t8 beat me to it, but I was really excited drawing it last night...

bd.pdf

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I didn't understand all of your answer, but I have a small remark. 6 inches is the hight of the hole's "wall", not neccesariy the diameter of the sphere. Actually, if the size of the hole was zero, the sphere would be 6 inches diameter.

Hope I uploaded my sketch correctly. I know t8t8t8 beat me to it, but I was really excited drawing it last night...

Maybe I messed up, I don't know. You may have been confused about sphere volume thing, look up Archimedes on wikipedia. I think it was volume, but maybe it was surface area. I got the impression that the hole is drilled 6 inches into the sphere, and the 'walls' of a cylindrical hole are curved.

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There are simpler ways to do this.

Just take the sphere with the hole in it and drop it inside a recipient full of water. Then take it out, and see how much water is missing. Phisics always helps

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This is my first post..

So I invite myself in...

The answer is the volume of a six inch sphere, which is 27 pi sq inches. (radius of 3, diameter of 6)

Moving to why:- when you drill from the top, whatever the size of the drill, the height of the cylinder so formed will reduce since the top cap would now be missing.

So to get a cylinder (or hole) of exactly six inches high, you must use a drill that is sufficiently large enough to leave only a small part of the sphere which is similar to a Ring so that the total height is exactly six inches.

the volume so formed is always equal to the volume of a six inch sphere.

I once heard a puzzle of having a steel wire that ties around the earth, then we add one meter of wire to it. how much would the earlier wire rise above the ground.

The answer is that it is all standard, irrespective whether it is an orange or the Earth.

:-)

Edited by InvincibleRam
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How can you possibly know the answer if you don't know how wide the whole is?

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How can you possibly know the answer if you don't know how wide the whole is?

That's what makes this a great puzzle.

Assume that all the needed information has been given.

The diameter of the sphere is not given.

Therefore the answer is the same for any sized sphere.

Therefore we can say the sphere diameter is 6"

A 6" hole through a 6" sphere has zero diameter, removing 0 volume from the sphere.

The remaining volume is the initial volume = 36 pi cubic inches.

If needed information is not given, then the answer is unknown.

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How can you possibly know the answer if you don't know how wide the whole is?

Sure you do. Its radius is (R^2-9)^(1/2) where R is the radius of the sphere.

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How can you possibly know the answer if you don't know how wide the whole is?

Sure you do. Its radius is (R^2-9)^(1/2) where R is the radius of the sphere.

Or if by "whole" you mean the entire sphere then its 2R.

Edited by maurice
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After reading through this post, I don't understand how this can be 36pi for the volume. Let's say the sphere is 6 inches in diameter, and you pull out a drill that has the same diameter as the sphere and somehow magically you drill through it... now your sphere is just wood chips on the floor.... How much volume is remaining? 0, because you made the whole sphere a hole...

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After reading through this post, I don't understand how this can be 36pi for the volume. Let's say the sphere is 6 inches in diameter, and you pull out a drill that has the same diameter as the sphere and somehow magically you drill through it... now your sphere is just wood chips on the floor.... How much volume is remaining? 0, because you made the whole sphere a hole...

Six inches in the Length, not the Diameter, of the hole.

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Confusion comes about as it is not exactly defined what the 6 inches drilled through is a measure of

I think this might explain what bonanova is asking

Lets just say for example that the sphere was the size of the earth the drill bit diameter is of such a width that by the time the earth has been drill through you have a doughnut 6 inches high. If this is the earth really thin sides with a massive hole in the middle and the drill has drilled a lot more than six inches. If the sphere is 6 inches high the drill bit must be 0 inches in diameter for the sphere to remain 6 inches high.

in other words this ( ) is 6 inches this O is not necessarily 6 inches

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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.

Quoting the original post, it says a 6 inch long hole, it doesn't specify the width.

I assume since it says 'through' the sphere, the hole must connect two outer edges of the sphere, be it straight through the center or off-centered or on a diagonal.

With only this information, I could theoretically get a drill bit whose diameter is the same as the sphere's diameter (or larger) and drill the sphere to nothingness no matter where the 6 inch long hole was drilled through it.

Google 'World's largest drill bit'. That would make a BIG difference compared to your standard hand held drill.

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Quoting the original post, it says a 6 inch long hole, it doesn't specify the width.

I assume since it says 'through' the sphere, the hole must connect two outer edges of the sphere, be it straight through the center or off-centered or on a diagonal.

With only this information, I could theoretically get a drill bit whose diameter is the same as the sphere's diameter (or larger) and drill the sphere to nothingness no matter where the 6 inch long hole was drilled through it.

Google 'World's largest drill bit'. That would make a BIG difference compared to your standard hand held drill.

If you had a drill bit whose diameter was the same as that of the sphere you would not be able to drill a 6 inch hole...so no you can't have the diameter be the same. The width of the hole is not specified but it is implied as the only possible radius of the hole that will generate a 6 inch hole is (R^2-9)^(1/2) where R is the radius of the sphere.

Edited by maurice
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With only this information, I could theoretically get a drill bit whose diameter is the same as the sphere's diameter (or larger) and drill the sphere to nothingness no matter where the 6 inch long hole was drilled through it.

If it was a 6 inch high cylinder you might have better luck. You would still have to successfully argue that nothing constitutes a 6 inch hole with infinitly thin sides. As it is a sphere if the drill is the same size as the sphere the hole at the end of drilling is 0 inches (the drill bit may have traveled 6 inches or more but we are measuring the hole not drill travelling). The depth of the hole (or height of the resulting doughnut) changes when you drill through the sphere as the drill knocks off the top and bottom of the sphere.

Although we are not told the width of the sphere or the width of the drill the width of sphere and drill are related. With a sphere with a 6 inch diameter the drill must be 0 inches wide, with a 60 inch diameter sphere the drill will need to be about 59.7 inches (the drill will travel 60 inches, but the remaining hole will be 6 inches)

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I suppose to sink the sphere in water after drilling, it will remove water equal to its volume.

Edited by wolfgang
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Hole in a sphere has been a lot of fun for a long tim. But has anyone yet addressed the question regarding the minimum length of the drill?

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