Volume of cube with three holes

[Zamaster]

I'm sorry but I looked this problem up on google, because I remember this problem being thrown at me by a tyrannical 8th grade algebra teacher trying to show what kind of problems 'he thought were difficult.' Now, I've taken a vector calculus class I noticed the problem is really easy -> if you take it from a geometric calculus standpoint. I say this because as soon as I learned the general domain version of a double integral, an analytic solution is obvious, as i simply wrote it out on paper while my professor was talking.

It comes out to be, given d the diameter of the hole and l the side length of a cube, l^3 - (3/4)*pi*l*d^2 + 2^(1/2)*d^3. Its quite elegant really. And you get this with four integrals, that is to say using other method appear to give way overblown answers. I'll only post my work if anyone near this (should be) dead topic is interested enough.

 

[Zamaster]

Haha reading this thread, yes, that's how I got the answer

Cube - ThreeCylinders + WeirdCircledCylinderThing + EvenWeirderCircledDoubleCylinderThing

 

[Bill Crean]

Zamaster's formula seems promising, but Zamaster does not state whether the formula is for the volume of the material removed from the cube or the volume remaining. For the case L=3 and D=1, we can easily calculate the volume of 6 cylinders with diametre = 1 and height = 1. It is of course 6*pi*D^2*L/4, which is, for D=1 and L(height)=1, 4.71238898 cubic (A) units. This is already greater than the value given by zamaster's formula and I have yet to include the intersecting part in the centre of the 3*3*3 cube. See below

L^3 - (3/4)*pi*L*D^2 + 2^0.5*D^3 = 9 - 7.0685833471 + 1.414213562 = 3.345630215.

The intersecting 3 cylinders, which occurs within the centre cube of the 3*3*3 occupies a volume (according to my calculations) of about 0.94 cubic units (B). Hence, the total volume removed from a 3*3*3 cube when a hole of diametre 1 unit is drilled through the centre of 3 orthogonal faces is A + B = 5.65 cubic units (2 decimal places).

The method I used was a non-calculus method and it was how the ancient Greeks would have solved it, if they were given the problem (namely section until a primative shape is established). I have a formula for my solution and it is given by:

3*pi*r*r(L-2r) + 8*r^3(2^0.5 - 1/2) + 12*r^3(pi/4 - 2^-0.5)(1 - 2^-0.5) - E (.05*r^3)

where r is the radius of the hole, L is the length of an edge of the cube, and E is an error term, which can be calculated depending how many decimal places of accuracy is required.

I stated Zamaster's solution looked promising: it might be a coincidence but 9 minus my answer (5.65) = 3.35, which is close to Zamaster's solution for the L=3, D=1 case.

 

[Bill Crean]

Did anyone spot my error - L^3 = 27, and so Zamaster's formula gives 21.34563009 (= 21.35 to 2 dec. places) for L=3 and D=1. This means that the Zamaster's formula yields the volume remaining after the holes are drilled, because my formula yields the volume removed namely 5.65 (2 dec. places) i.e. 27 - 5.65 = 21.35.

This means the problem has been solved successfully by two different methods.

 

[Zamaster]

Volume remaining! The fact that it comes out to square root of two boggles my mind! Yeah, my solution is entirely analytical without any approximations, but yes I did use calculus.

 

[zgozvrm]

I may be repeating someone here, but I scanned the above posts and didn't see any reference to this:

It may be easier to think of the cube kind of like a Rubik's cube. Cut it into sections such that only a 1/2" cube remains in the center. Determine the volume of the removed parts first. Then work on the center piece separately.

 

[Zamaster]

I mean, we have two solutions, one analytical using calculus, and one geometric numerical. I think this problem has been taken care of? Unless there are other cool methods to solve it.

 

[Bill Crean]

There is nothing wrong with using calculus, of course. When I was given the problem (40 years ago) the statement of the problem indicated no advance maths. or computer programs to be used (there were no PCs then in any case). I think that the calculus method must be much shorter than mine. While the error term in my formula allows me to calculate the solution to any required accuracy, I think it removed elegance to the sectioning method, which could only get the right answer to 2 decimal places.

The person, incidentally, who gave me the problem did not have the solution. I wonder if he ever solved it.

 

[Bill Crean]

Yes, there is another method of solving the problem, as given by a silly old engineering colleague: Fill the holes in the cube with a liquid and then pour the liquid into a measuring cylinder!