# hole in a cylinder

### #1

Posted 17 January 2008 - 03:45 AM

A six inch high cylindrical hole is drilled through the center of a sphere. How much volume is left in the sphere?

### #2

Posted 17 January 2008 - 10:46 AM

where:

rs = radius sphere

rc = radius cylinder

h = 6 inches

ie - volume of sphere - volume of cylinder.

Then simplify.

### #3

Posted 17 January 2008 - 01:35 PM

### #4

Posted 17 January 2008 - 02:47 PM

There is a unique answer, identical for any size sphere.

That's doubtful. We know neither the size of the sphere, nor the radius of the cylinder. Are you sure you're not forgetting something?

Example: drill a 6 inch hole with radius 6 inches through the center of a six inch diameter sphere. Zero volume is left.

### #5

Posted 17 January 2008 - 07:05 PM

I read the book so I won't tell you what the answer is but I can assure you it exists and is unique.

### #6

Posted 17 January 2008 - 07:10 PM

Technically as it`s Open, the Volume would be EVERYTHING EVERYWHERE in the universe, Or it would remain exactly the same if you Imagined it was still the same shape.

your question is VERY vague!

### #7

Posted 17 January 2008 - 08:14 PM

- The question is a typical 1st semester physics homework, i.e. not some super-complicated problem requiring advanced math.

- It is not obvious (to me) that there is a unique answer. But thinking about the problem a bit, it's also not obvious that there shouldn't be one. Making up examples with spheres with a diameter less than 6 inches just looks like an attempt to cover the own inability by blaming someone else of not being sufficiently specific. If three different people (the book as original source also counted) tell you there is a unique answer, then perhaps that simply is the case.

-- As a matter of fact, if you take "there is a unique answer regardless of the sphere size" as granted (you shouldn't, you'll miss the point of the exercise), then you can trivially get the answer (which is 36pi inch³, btw) from taking a sphere with a radius of 3 inch (because the hole then has -asymptotically, if you want- a volume of 0).

- The point of the problem is that two of the unknowns (diameter of the hole and radius of the sphere) cancel in such a way, that the result is just a plain number. If you just hope for that and straightforwardly calculate the volume, then the problem is solved by 1 sketch, two simple relations between the appearing variables and a 5-line calculation. Hardly unsolvable compared to that typical topics on sfn are about quantum gravity or climate models.

### #8

Posted 17 January 2008 - 08:29 PM

but if you Imagine it`s a still a sphere then the volume stays the same, it doesn`t matter what it`s made of inside, where there`s a 6dia bore full of air and the rest is solid plastic, it`s volume inside is the same.

there`s no need for ANY numbers or maths at all

### #9

Posted 17 January 2008 - 09:17 PM

Perhaps you can tell me what part of the problem forbids my solution?

### #10

Posted 17 January 2008 - 10:15 PM

For example that you cannot drill a hole with a diameter of 12 inch into a sphere with a diameter of 6 inch.Perhaps you can tell me what part of the problem forbids my solution?

EDIT: A for a picture visualizing a sphere and a sphere with a cylindrical hole in it:

http://www.sciencefo...=1&d=1200608334

Left greyish object is a sphere, the right one is a sphere with a cylindrical hole in it.

### #11

Posted 17 January 2008 - 10:24 PM

I take it that if the hole is 6 inches deep and goes all the way through, then there are a limited set of solutions- one for each sphere. As the sphere gets larger, the edges of he whole move nearer (in terms of angle not necessarily size) to each other.

If we take a calculus based approach here, a sphere of radius r is made up of a series of infinitessimal disks, but when the cylindrical hole is drilled, it becomes a series of infinitessimal annuli. moving upwards from the "equator" with h increasing , the radius, y, of any one disk is given by

Taking a cylindrical hole of radius x out with the cross section of the cylinder to what I previously called the "equator", the width on an annulus w is,

And when the value of h is equal to 3 (i.e. the limit of a 6 inch hole, half being each side of the equator) the width of the annulus is 0, thus,

Which gives

And the volume of an annular element is thus given by

Summing for the hole depth of 6 inches,

### #12

Posted 17 January 2008 - 10:44 PM

- I personally disagree with "only one solution for a given sphere", but that's really just semantics. For me, two different orientations of the drilling are two different solutions, but of course they will result in the same remaining volume.

- The edges not only "not necessarily" move closer in terms of absolute distance, they don't do that at all -> the distance is always 6 inches, of course.

- You forgot a factor of pi in the term for the area of a circle, meaning your final result is off exactly by this factor.

EDIT: @ post #13. No, it was pretty clear what you meant. I just thought that writing "you're off by a factor of pi" was too little for a post

### #13

Posted 17 January 2008 - 11:00 PM

Might as well get your moneys worth;)

### #14

Posted 18 January 2008 - 03:11 PM

### #15

Posted 19 January 2008 - 05:21 PM

And, for Mr skeptic's benefit

"Perhaps you can tell me what part of the problem forbids my solution?"

The fact that a cylinder of zero height is not a cylinder with a height of six inches as specified in the question. (I'm sure I mentioned that before.)

### #16

Posted 21 January 2008 - 03:13 AM

Despite what John Cuthber says, in my example the cylinder is 6 inches deep (not zero), and 6 inches wide, so that the entire sphere would be contained in the cylinder. If that seems wrong to you, imagine a sphere infinitesimally larger than 6 in diameter, so that there would be an arbitrarily small ring and endcaps left. If there wasn't a restriction on the radius of the cylinder, the answer would be arbitrary.

Of course, it is much clearer now that ishmael explained in post 14.

Then, for a sphere of radius R, the radius r of the cylinder will be

(where 3 inches is half the height of the cylinder)

Actually, that is the same as MrMongoose said. At first I thought his was for a specific sphere, but he has the general solution, so I won't redo it.

### #17

Posted 21 January 2008 - 08:01 PM

What I think should have been added to the original question, however, is that the hole goes through the sphere from one surface to another. I just assumed that that was what was meant as there was apparently a unique solution.

### #18

Posted 21 January 2008 - 08:27 PM

If you put a drill through something at least one of the "endcaps" is removed; generally both.

Without the endcaps what you have is an arbitrarily small ring which isn't six inches high.

I accept it's a sloppy definition, but I'd usually think that if someone says they put a cylinder through a point (like the centre of a sphere) they meant the axis of the cylinder went through the point. The reason I'd think that was that otherwise, it doesn't really tell you much.

### #19

Posted 21 January 2008 - 08:33 PM

I think so, too. Espeically in this case: If the axis of the hole does not go through the center of the sphere, you don't end up with a cylidrical hole.I accept it's a sloppy definition, but I'd usually think that if someone says they put a cylinder through a point (like the centre of a sphere) they meant the axis of the cylinder went through the point.

### #20

Posted 22 January 2008 - 07:23 PM

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