Hole and sphere.

[mfgoode]

A cylindrica hole six inches long has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?

Mally

[retina]

I remember seeing this puzzle years ago and was surprised to discover that (IIRC) the radius of the sphere is completely cancelled out and the volume is the same no matter how big we make the radius.

But I think the question should be posed a little differently to avoid confusion. Perhaps something like "What is the volume of the remainder in the sphere"?

[mfgoode]

Well Retina, suit yourself as to the presentation of the problem. It was first posed by the puzzilist Sam Loyd and adapted in several other publications.

Well , at least you did not comment that it has insufficient data to be solved.

So what is the the volume of the rest of the sphere after the hole has been bored?

Hint: Dont go into calculus and the rest. There is also a purely logical solution too! Without that, one must at least solve it by pure geometry though you must arrive at the volume of the caps which requires a formula not too well known.

Mally

P.S. I'm sorry I did not take note of your spoiler (though I read it} and concentrated on the re-wording of the problem.
You are quite right, though.

[xilman]

Originally Posted by mfgoode

A cylindrica hole six inches long has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?

Mally

This is a classic.
As the question doesn't mention the diameter of a sphere, the answer must be independent of the size of the sphere, provided that the hole can be six inches long. Therefore, chose a sphere of six inches diameter. The hole must then be of zero diameter and so of zero volume. Consequently, the answer is the same as the volume of a six inch diameter sphere. The volume of a sphere of radius r is 4\pi r^3/3. Therefore the answer to the question is 4\pi 3^3/3 cubic inches, or 36\pi cubic inches.

Paul

[akruppa]

We had this puzzle some time ago: http://www.mersenneforum.org/showthread.php?t=5460

Alex

[Orgasmic Troll]

Originally Posted by xilman

This is a classic.
As the question doesn't mention the diameter of a sphere, the answer must be independent of the size of the sphere, provided that the hole can be six inches long.

Paul

This assumes that the question is well posed. Considering the source, I wouldn't be comfortable making this logical leap had I not seen the problem before

[mfgoode]

Originally Posted by Orgasmic Troll

This assumes that the question is well posed. Considering the source, I wouldn't be comfortable making this logical leap had I not seen the problem before

You don't have too OT.

Here is a more rigorous solution by an analogy in two dimensions. Take a cross section of the 'completed' residue of the sphere. You will get two concentric circles. Let us call it a circular track. Pl. draw a figure for a better view.
Now given the longest straight line that can be drawn on the circular track of ANY dimensions. This is a tangent to the smaller circle.

The area of the track will be equal to the area of a circle having the straight line as a diameter. No assumptions here as it can be proved! Convince your self if you like by proving it. It should be called a theorem!

Now reduce the dia. of the central smaller circle. The straight line (Tangent) will become longer. In the limit the straight line becomes the dia. and the area of the smaller circle zero.
We are given the depth of the hole is 6 inches and so convert the problem from area of the cross section to volume of the sphere of dia. 6 inches.
And this is 36 pi. a constant. In other words the residue is constant regardless of the hole's diameter or the size of the sphere. The diameter of the sphere will be the same as the depth of the hole. Vary these if you like to make it general.

Mally

ATTN: Akruppa: I doubt if this view was given in the last time this problem was presented in '06

[Wacky]

Originally Posted by mfgoode

And this is 36 pi. a constant. In other words the residue is constant regardless of the hole's diameter or the size of the sphere.

The diameter of the sphere will be the same as the depth of the hole.

Actually, this last portion of your statement is incorrect. The diameter of the sphere is greater than (or in the limiting case equal to) the depth of the hole. The hole removes a right cylindrical plug whose height is the depth of the hole PLUS two "end caps". But, as noted, the residue is of a constant volume even though the diameter of the sphere varies.

[mfgoode]

Originally Posted by Wacky

Actually, this last portion of your statement is incorrect. The diameter of the sphere is greater than (or in the limiting case equal to) the depth of the hole. The hole removes a right cylindrical plug whose height is the depth of the hole PLUS two "end caps". But, as noted, the residue is of a constant volume even though the diameter of the sphere varies.

Thank you Wacky! I stand corrected and an astute observation on your part.
I welcome this type of correction instead of posters taking digs at presentation or mis spelling or such trivia which breeds unpleasantness.

Mally

[davieddy]

By additionally considering the intersection of
a plane tangential to the cylinder with the sphere (a circle) I can deduce
from Mally's theorem that the largest area of Mally's cross section is
that of this circle (9PI) whatever the diameter of the sphere.
However it is far from obvious that the area of other rings stacked in
the 6 inch height are similarly independent of the sphere's diameter.

Am I missing something?

David

[davieddy]

Let the radius of the sphere be R inches.
Take its intersection with a plane h inches from
the centre of the sphere, and let the radius of this
circle be r inches.
r^2=R^2 - h^2
The square of the radius of the cylinder is (R^2 - 3^2) inches^2

To find the area of the ring ("track" as Mally called it) we
subtract these values (*PI).

Lo and behold, R leaves the problem.

I think Mally half showed this for h=0.

David