Volume of a Sphere

[Patrick123]

This is quite an old puzzle, but I don't think it's been here before.

A cylindrical hole 6 cm long is drilled straight through the centre of a sphere. What is the volume remaining in the sphere?

Patrick

[drew]

Originally Posted by Patrick123

This is quite an old puzzle, but I don't think it's been here before.

A cylindrical hole 6 cm long is drilled straight through the centre of a sphere. What is the volume remaining in the sphere?

Patrick

What is the radius of the hole, and/or of the sphere?

Is the length of the hole measured at the surface of the cylinder, or its axis?

Drew

[Patrick123]

Originally Posted by drew

What is the radius of the hole, and/or of the sphere?

Is the length of the hole measured at the surface of the cylinder, or its axis?

Drew

That is the beauty of this question. All you have is the length of the hole...

Patrick

[drew]

Originally Posted by Patrick123

That is the beauty of this question. All you have is the length of the hole...

Patrick

Zero. The hole is 6" measured thrugh its axis, and has a radius of 3". The sphere is 6" in diameter.

[Patrick123]

Originally Posted by drew

Zero. The hole is 6" measured thrugh its axis, and has a radius of 3". The sphere is 6" in diameter.

Nope. That is not the answer.

[drew]

Originally Posted by Patrick123

Nope. That is not the answer.

36*pi in[sup]3[/sup]

[Patrick123]

Correct Drew. I love this problem for the seemingly sparse information that you're given.

[drew]

Originally Posted by Patrick123

Correct Drew. I love this problem for the seemingly sparse information that you're given.

I know I was being difficult, but your first response clued me in. To be honest, I guessed at the answer assuming an infinitessimal 'filament' for a hole (in other words, it's just a volume of a sphere of 6" diameter).

Now that I've looked further, I'd like to propose what I think is a very elegant solution.

You can think of the volume of the remainder of the sphere as an integration of cross-sections over the height of the sphere. So what do we know?

Assuming the center of the sphere is at height 0, the rim of the sphere will be a height 3" from the center, and will be a distance 'R' from the center...the radius of the sphere.

Using the pythagorean theorem, at an arbitrary height, 'h' the radius of the sphere's cross-section is sqrt(R² - h²). The radius of the hole is invariably the radius of the sphere's cross-section at h=3, which is sqrt(R² - 3²)

Since the area of the ring that represents the cross-section is a difference of squares of these values, the Radius cancels out:

A(h) = pi * [(R² - h²) - (R² - 3²)]

= pi * (9 - h²)

which is a function of neither the sphere's or cylinder's radius. Therefore, without going through the trouble of integrating the above, it's reasonable to assume the answer for R=3 (volume of a regular sphere) applies to all cases.

But the integral's easy:

9*pi*h - 1/3*pi*h³

evaluated from -3 to 3:

9*pi*6 - 1/3*pi*(27 - -27)

= 54*pi(1 - 1/3)

= 36*pi

Another great puzzle!

[Patrick123]

Originally Posted by drew

I know I was being difficult, but your first response clued me in. To be honest, I guessed at the answer assuming an infinitessimal 'filament' for a hole (in other words, it's just a volume of a sphere of 6" diameter).

Now that I've looked further, I'd like to propose what I think is a very elegant solution.

You are right on both methods.
The first, as the width of the hole approaches zero, the height of the cylinder reaches 6"
and the second, subtracting the volume of the cylinder and the volume of the two caps will give
you an equation where the unknowns cancel out leaving 36pi.

Patrick

[mfgoode]

Originally Posted by Patrick123

You are right on both methods.
The first, as the width of the hole approaches zero, the height of the cylinder reaches 6"
and the second, subtracting the volume of the cylinder and the volume of the two caps will give
you an equation where the unknowns cancel out leaving 36pi.

Patrick

Note how adroitly this shows that the radius of the sphere a and radius of the hole b appear only as (a^2 - b^2) which is equal to the square of half the depth of hole, regardess of the value of a or b.
Apparently, there is no other 'physical explanation' of the surprising fact that all 6 inch long holes drilled in all spheres 6 inches or bigger leave the same residue. :surprised
Mally

[Patrick123]

Originally Posted by mfgoode

Note how adroitly this shows that the radius of the sphere a and radius of the hole b appear only as (a^2 - b^2) which is equal to the square of half the depth of hole, regardess of the value of a or b.
Apparently, there is no other 'physical explanation' of the surprising fact that all 6 inch long holes drilled in all spheres 6 inches or bigger leave the same residue. :surprised
Mally

It is fascinating, this is how I originally worked it out.

Let R = radius of sphere.

Radius of hole :=sqrt( R^2 -9).
The altitude(A) of the spherical caps that would be removed from each end will be R - 3.
The volume of the cylinder(hole) will be 6*pi*(R^2-9).
The volume of the spherical cap will be pi*A*(3*R^2+A^2) - 6.
The volume of the sphere is 4*pi*(R^3-3).

We now have: Remaining Volume = Volume of sphere - ( Volume of cylinder + 2* Volume of cap).

All the terms cancel out and we are left with 36pi.