Sphere with a hole through it

[Gengulphus]

An old maths puzzle that some of you may well have encountered before, but I think it worth posting for those who haven't:

I have a large wooden sphere. I drill a cylindrical hole through it, centrally positioned (i.e. the axis of the hole passes through the centre of the sphere). Measuring the hole, I find that it is 6 inches long.

What is the volume of the piece of wood I'm left with?

Gengulphus

[GoSeigen]

An old maths puzzle that some of you may well have encountered before, but I think it worth posting for those who haven't:

I have a large wooden sphere. I drill a cylindrical hole through it, centrally positioned (i.e. the axis of the hole passes through the centre of the sphere). Measuring the hole, I find that it is 6 inches long.

What is the volume of the piece of wood I'm left with?

Gengulphus

Spoiler
I feel like I am missing something here, but the question implies that the volume is independent of the diameter of the hole. I suppose I should prove it, but if the implication is fair, then the question is trivial and the volume of the piece of wood is (4/3)pi*6^3, determined by assuming an infinitely thin drill. When I have time I'll work on the proof.

[servodude]

Spoiler

- i think you might have forgotten to divide by 2 somewhere

I had a real life experience of this ilk at Glasgow airport once.
My 6 year old wasn't initially allowed a snow globe she'd been given by a friend in her carry on because of the 100ml rule
- I asked them to show me their working... and they let us take it

- sd

[GoSeigen]

Spoiler

- i think you might have forgotten to divide by 2 somewhere

Oops!

I was more focussed on the gross liberty I was taking with the question...


GS

[UncleEbenezer]

I'm not going to try and draw this. You'll have to visualise the angle θ from the central plane to the rim of the hole.

The volume of the bead is that of the sphere, minus a cylinder, minus two caps. Each cap is a sector minus a cone. That's some standard formulae, but first let's have some notation.

Let S be the radius of the sphere, and R that of the cylinder (and cones), and L the length of the cylinder (given as 6 inches). These are then related by R = S cosθ and L = 2 S sinθ .

Now we have the standard formulae:
Sphere = 4/3 π S^3
Cylinder = L π R^2
Sector = 2/3 π S^3 (1 - sinθ) (n.b. I had to look that one up)
Cone = π/3 R^2 (L/2)

So Bead = Sphere - Cylinder - 2 (Sector - Cone)
= 4/3 π S^3
- L π R^2
- 4/3 π S^3 (1 - sinθ)
+ π/3 R^2 L

Or subtracting the cones from the cylinder:
Bead = 4/3 π S^3
- 2/3 π R^2 L
- 4/3 π S^3 (1 - sinθ)

Expressing everything in terms of S and θ:

Bead = 4/3 π S^3
- 2/3 π (2 S sinθ) (S cosθ)^2
- 4/3 π S^3 (1 - sinθ)

= 4/3 π S^3 ( 1 - sinθ cosθ^2 - (1 - sinθ) )

Aha! That's
= 4/3 π S^3 sinθ(1 - cosθ^2)
and by Pythagoras
= 4/3 π (S sinθ)^3

But S sinθ is half the length of the hole, given as 6 inches. So that's 4/3 π * 27, or 36π cubic inches.

[edit] GoSeigen's colour works better than my original

[Gengulphus]

Well done both to GoSeigen for finding the essence of the 'maths puzzle' solution, which is based on the assumption that the answer doesn't depend on unspecified information, and to UncleEbenezer for finding a 'maths problem' solution that makes no such assumption!

I think I first encountered this puzzle in one of Martin Gardner's "Mathematical Games" columns in Scientific American, probably somewhere around 50 years ago.

Gengulphus

[mc2fool]

I must admit, my very first reaction to the puzzle was, "uh, he's forgotten to tell us the diameter of the hole", pretty quickly followed by, "ah, wait, this is from Gengulphus and he doesn't make that sort of mistake!", (well, not that I ever spot, anyway!) and so, like GoSeigen, figured that it wasn't relevant...

... and I then got boggled on the very idea that the volume of the remaining wood would be the same for a wooden sphere you could hold in your hand with a very thin 6" hole drilled in it and one the size of a planet that had almost all of it removed leaving just an extremely thin six inch high ring tens of thousands of kilometers across! I still am.

I suspected the proof would be beyond me, although GoSeigen's solution is delightfully lateral and UncleEbenezer's solution I think I could deal with if I dusted off the long-unused O-level trig section of my brain.

Anyway, thinking I couldn't do it I googled and I did find another solution (that is beyond me; calculus, shudder!) that might be of interest to those interested in such things! http://www.sfu.ca/~adebened/funstuff/sphere_cyl.pdf

[UncleEbenezer]

Well done both to GoSeigen for finding the essence of the 'maths puzzle' solution, which is based on the assumption that the answer doesn't depend on unspecified information, and to UncleEbenezer for finding a 'maths problem' solution that makes no such assumption!

I think I first encountered this puzzle in one of Martin Gardner's "Mathematical Games" columns in Scientific American, probably somewhere around 50 years ago.

Gengulphus

While mine is a 'maths problem' solution, I'd argue GoSeigen's is the better mathematicians' solution.

The mathematician is (as my O/A level maths teacher always explained[1]) lazy, and doesn't do unnecessary work. Hence the true mathematician uses available information: in this case, that Gengulphus as puzzle-setter won't have cocked up at a level that invalidates his solution.

Having already been posted, that solution was not available to me, though I'd have preferred it over my own. Spoiler-hiders are good for reading, but should IMO be checked before posting what might easily be a duplicate of something so short&simple!

[1] Though the expectation to "show your working" was often less-than consistent with that.

[Gengulphus]

Well done both to GoSeigen for finding the essence of the 'maths puzzle' solution, which is based on the assumption that the answer doesn't depend on unspecified information, and to UncleEbenezer for finding a 'maths problem' solution that makes no such assumption!

I think I first encountered this puzzle in one of Martin Gardner's "Mathematical Games" columns in Scientific American, probably somewhere around 50 years ago.

While mine is a 'maths problem' solution, I'd argue GoSeigen's is the better mathematicians' solution.

The mathematician is (as my O/A level maths teacher always explained[1]) lazy, and doesn't do unnecessary work. Hence the true mathematician uses available information: in this case, that Gengulphus as puzzle-setter won't have cocked up at a level that invalidates his solution.

But the true mathematician also doesn't make assumptions without stating them, and there are at least two there: firstly, that I haven't deviated from that normal pattern of mine, and secondly, that I'm expecting an answer that is a number rather than a formula...

Gengulphus

[XFool]

Pardon me for butting in - not really my scene.

But I cannot follow the working out as given. In trying to follow I too had to look up the formula for the volume of a spherical sector, on Wikipedia.
It gave: 2/3 π S^3 (1 - cosθ)

https://en.wikipedia.org/wiki/Spherical_sector#Volume

ASSUMING θ is the angle between the centre line of the hole and the radius of the sphere, S, at the edge of the hole.

But if so, then I make R = S sinθ and L = 2 S cosθ

Where am I going wrong? (Perhaps it all works out the same?)

[Gengulphus]

Pardon me for butting in - not really my scene.

But I cannot follow the working out as given. In trying to follow I too had to look up the formula for the volume of a spherical sector, on Wikipedia.
It gave: 2/3 π S^3 (1 - cosθ)

https://en.wikipedia.org/wiki/Spherical_sector#Volume

ASSUMING θ is the angle between the centre line of the hole and the radius of the sphere, S, at the edge of the hole.

But if so, then I make R = S sinθ and L = 2 S cosθ

Where am I going wrong? (Perhaps it all works out the same?)

The angle you're calling θ is "the angle between the centre line of the hole and the radius of the sphere, S, at the edge of the hole", which is called φ in the Wikipedia link. The angle UncleEbenezer is calling θ is "the angle θ from the central plane to the rim of the hole". I.e. you're measuring the angle from the 'polar axis' of the sphere, UncleEbenezer is measuring it from the 'equatorial plane' of the sphere. Using θ to denote UncleEbenezer's version and φ to denote yours, they're related by φ = π/2 - θ, so sinφ = cosθ and cosφ = sinθ.

Gengulphus

[XFool]

...OK. Thanks.

[Gengulphus]

I suspected the proof would be beyond me, although GoSeigen's solution is delightfully lateral and UncleEbenezer's solution I think I could deal with if I dusted off the long-unused O-level trig section of my brain.

Anyway, thinking I couldn't do it I googled and I did find another solution (that is beyond me; calculus, shudder!) that might be of interest to those interested in such things! http://www.sfu.ca/~adebened/funstuff/sphere_cyl.pdf

Yes, that solution with its triple integral and polar co-ordinates is a bit intimidating to the calculus-allergic! However, two of the integrals and the polar co-ordinates aren't necessary, so here's a spoiler-protected solution that uses calculus pretty minimally, and as a bonus no trigonometry other than Pythagoras' theorem:

Let R be the radius of the sphere and r the radius of the cylinder. Looking at a cross-section containing the axis of the cylinder, we can see a right-angled triangle made up from the radius of the sphere from its centre to the rim of the hole, the radius of the cylinder to halfway through the hole, and half the length of the hole. So by Pythagoras' theorem, R^2 = r^2 + (6/2)^2 = r^2 + 9.

Now look at cross-sections perpendicular to the axis of the cylinder. Of those cross-sections, one passes through the centre of the sphere, and it intersects the remaining piece of wood in an annulus of inner radius r and outer radius R. That annulus has area πR^2 - πr^2 = 9π. Now consider a cross-section a distance h (which is <= 3) away from that cross-section: its intersection with the remaining wood is still an annulus with inner radius r, but by another application of Pythagoras' theorem, its outer radius is sqrt(R^2-h^2), and that annulus has area π(sqrt(R^2-h^2))^2 - πr^2 = π(R^2-h^2) - πr^2 = πR^2 - πh^2 - πr^2 = π(9-h^2). Now the one necessary bit of calculus, which is knowing that the volume of an object is the integral of the areas of parallel cross-sections through it. That tells us that the volume of the remaining piece of wood is twice the integral from h=0 to h=3 of π(9-h^2).

We can then observe that that integral doesn't depend on R, mathematically justifying the assumption in GoSeigen's solution. (Informally, we've shown that the cross-sectional area of the remaining piece of wood at every height above the 'equatorial plane' (or depth below it) is identical to the corresponding cross-sectional area of a sphere of radius 3, and that's enough to tell us that the volumes of the remaining piece of wood and of that sphere are identical.)

Or if one isn't too calculus-allergic, one can apply a bit more calculus, namely the knowledge that integration behaves linearly and that the indefinite integral of h^n is (h^(n+1))/(n+1), to say that the indefinite integral of π(9-h^2) is 9π*(h^1)/1 - π*(h^3)/3 = 9πh - (πh^3)/3. The definite integral from h=0 to h=3 is that formula evaluated at h=3 minus the same formula evaluated at h=0, which is (27π - (27π)/3) - (0π - (0π)/3) = 18π, and the volume of the remaining piece of wood is twice that, or 36π.

Gengulphus

[ChrisNix]

An old maths puzzle that some of you may well have encountered before, but I think it worth posting for those who haven't:

I have a large wooden sphere. I drill a cylindrical hole through it, centrally positioned (i.e. the axis of the hole passes through the centre of the sphere). Measuring the hole, I find that it is 6 inches long.

What is the volume of the piece of wood I'm left with?

Gengulphus

Spoiler
I feel like I am missing something here, but the question implies that the volume is independent of the diameter of the hole. I suppose I should prove it, but if the implication is fair, then the question is trivial and the volume of the piece of wood is (4/3)pi*6^3, determined by assuming an infinitely thin drill. When I have time I'll work on the proof.

Half marks.