Three intersecting cylinders

[bonanova]

A classic problem solved by Archimedes before the invention of calculus, concerns two circular cylinders intersecting at right angles.

If each cylinder has a radius of 1 unit, what is the volume of space that is common to the cylinders?

For warm-up, you can figure out the answer - using calculus if you like.

Here's the puzzle of the day:

What is the volume of space that is common to three orthogonally intersecting cylinders of unit radius?

In both problems the axes of the cylinders intersect.

[Pickett]

I've done the 2 cyclinder problem before...so that really helps with starting this problem. I'll leave that one to someone else...

I beleive this can be represented by this integral (this could be very wrong, as it's been a while...):

16 * ₀∫^pi/4 ₀∫¹ ₀∫^{√(1 - r^2cos^2Θ)} r dz dr dΘ

So, solve the inner integral:

16 * ₀∫^pi/4 ₀∫¹ r * √(1 - r²cos²Θ)dr dΘ

Then the next integral (this is the one I'm really not sure I did right):

-16/3 * ₀∫^pi/4 ((sin³Θ - 1) / cos²Θ) dΘ

Finally, the last integral:

16 - 8√2...or approx 4.69

[bonanova]

I've done the 2 cyclinder problem before...so that really helps with starting this problem. I'll leave that one to someone else...

I beleive this can be represented by this integral (this could be very wrong, as it's been a while...):

16 * ₀∫^pi/4 ₀∫¹ ₀∫^{√(1 - r^2cos^2Θ)} r dz dr dΘ

So, solve the inner integral:

16 * ₀∫^pi/4 ₀∫¹ r * √(1 - r²cos²Θ)dr dΘ

Then the next integral (this is the one I'm really not sure I did right):

-16/3 * ₀∫^pi/4 ((sin³Θ - 1) / cos²Θ) dΘ

Finally, the last integral:

16 - 8√2...or approx 4.69

Bravo, for accuracy and beautiful symbols. Two coveted stars!

[kbrdsk]

Is the three cylinder problem also doable without using calculus?

[bonanova]

Is the three cylinder problem also doable without using calculus?

Well to be fair, Cavalieri's principle was the calculus dodge for the two cylinder calculation.

It's a near cousin to the integral calculus.

The three cylinder problem might succumb in a similar way.