## Recommended Posts

This is from an old Martin Gardner book.

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

##### Share on other sites

V = (4/3 Pi.rs^3)sphere - (Pi.rc^2 . h)cylinder

where:

h = 6 inches

ie - volume of sphere - volume of cylinder.

Then simplify.

##### Share on other sites

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

##### Share on other sites
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.

##### Share on other sites

Since the length of the cylinder is specified as six inches "A six inch high cylindrical hole" your idea, a cylinder of zero length, is excluded so it's not a valid criticism.

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

##### Share on other sites

you never said how big the sphere was at the outset, and more to the point Would it still Be a sphere with a hole in it?

Technically as its 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.

##### Share on other sites

- 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.

##### Share on other sites

but thats just it, if you drill a hole through it, its no longer a Sphere! its more like a bead you thread on a string.

but if you Imagine its a still a sphere then the volume stays the same, it doesnt matter what its made of inside, where theres a 6dia bore full of air and the rest is solid plastic, its volume inside is the same.

theres no need for ANY numbers or maths at all ##### Share on other sites

The reason I gave my example is because it is a valid solution, and shows that whoever said the problem gave the same answer for every sphere is wrong. Unless there is a restriction or relation between the size of the sphere and the radius of the cylinder, than I can choose any I please. Was there a picture that further restricted the problem?

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

##### Share on other sites
Perhaps you can tell me what part of the problem forbids my solution?

For example that you cannot drill a hole with a diameter of 12 inch into a sphere with a diameter of 6 inch.

EDIT: A for a picture visualizing a sphere and a sphere with a cylindrical hole in it: Left greyish object is a sphere, the right one is a sphere with a cylindrical hole in it. ##### Share on other sites

I think this is right... 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

$y=(r^2 -h^2)^\frac{1}{2}$

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,

$w=(r^2 -h^2)^\frac{1}{2} -x$

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,

$0=(r^2 -3^2)^\frac{1}{2} -x$

Which gives

$x=(r^2 -9)^\frac{1}{2}$

And the volume of an annular element is thus given by

$\frac{dV}{dh}=\pi((r^2 -h^2)^\frac{1}{2})^2-\pi((r^2 -9)^\frac{1}{2})^2$

$\frac{dV}{dh}=\pi(r^2 -h^2-r^2 +9)$

$\frac{dV}{dh}=\pi(9 -h^2)$

Summing for the hole depth of 6 inches,

$V=\pi\int_{-3}^{3}(9 -h^2)dh$

$V=\pi[9h-\frac{h^3}{3}]^3_-3$

$V=\pi(27-9)-\pi(-27+9)=36\pi inches^3$

##### Share on other sites

- 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 ##### Share on other sites

Can't argue with the third one so I'll go about fixing it, and I agree the first thing is a bit shoddy. As for the second, I said they move closer in terms of angle, i.e. the cylinders length increases with respect to its radius. Again perhaps I should have been clearer!

Might as well get your moneys worth;)

##### Share on other sites

Perhaps I should have stated that the sphere must be at least 6" in diameter, but I believe that is implied in the original problem, as stated. What is amazing to me is that the answer is unique for all sizes of spheres, from 6" up to the size of the earth, and beyond. A six inch long cylindrical hole drilled through a sphere the size of the earth would be shaped as a 6" thick disc of approximately 8000 miles diameter. It would leave a very thin ring of material six inches wide and 8000 miles in diameter. This ring would have the same volume as a six inch diameter sphere with a cylindrical hole of zero diameter. Do the math. Amazing, no?

##### Share on other sites

I think that saying you drill a six inch long hole in it already says the shpere must have been more than six inches in diameter; explicitly saying so should be redundant. and thanks, btw to Mr Mongoose for doing the maths.

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.)

##### Share on other sites

As I was saying, the original problem in post #1 made no reference to the radius of the cylinder, nor put any restrictions on how the hole related to the sphere other than that it passed through the center. In particular, it never said the cylinder was centered where the sphere was centered, nor that the endcaps of the sphere would get discarded, nor that the radius had to be such that the circles of the cylinder are on the surface of the sphere. True, I could have guessed all this, but I prefer not to solve a problem if I'm not sure it is the right one.

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

$3^2 + r^2 = R^2$ (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.

##### Share on other sites

The point is that it isn't a 6" cylinder projected through a sphere and the overlap discarded. It is a 6 inch deep cylindrical hole in the sphere, so it MUST pass through the centre and its ends must be on the surface of the cylinder.

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.

##### Share on other sites

"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 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.

##### Share on other sites
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.

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.

##### Share on other sites

Oops! I should have spotted that too.

##### Share on other sites
The reason I gave my example is because it is a valid solution, and shows that whoever said the problem gave the same answer for every sphere is wrong. Unless there is a restriction or relation between the size of the sphere and the radius of the cylinder, than I can choose any I please. Was there a picture that further restricted the problem?

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

The hole in your sphere would not be 6" TALL....

##### Share on other sites

The name you'd give probably depends on the proportions.

http://www.scienceforums.net/forum/attachment.php?attachmentid=1711&stc=1&d=1201032698

I might call the right one a ring, but I'd rather not do that for the middle one.

EDIT: @Dog: 6 inches is exactly the size I'd expect from a 6 inch deep hole. That's the point of the question a lot of people do not seem to understand. ##### Share on other sites

I'd probably call the one in the middle a bead. The point is that if you were to thread the bead/ ring/whatever else you might want to call it onto a tight fitting string then it would hide six inches of that string.

## Create an account

Register a new account