Challenge Math Challenge by Charles Link #1

[Greg Bernhardt]

Submitted by @Charles Link
Solved by: @MAGNIBORO and @maline

RULES:
1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

CHALLENGE:
Beginning with a cube with each side of length 1", drill a 1" diameter hole all the way through in each of the 3 perpendicular directions. Find the remaining volume.

 

[Nidum]

Quite a fun shape to play with . Interesting question is whether it is now one body or eight bodies ?

 

[Charles Link]

I will provide one item of interest concerning the solution=it does have a closed form solution, and the final expression you get is considerably simpler than you might expect.

 

[MAGNIBORO]

nice problem, I'm going to download a software to help me visualize this.
i was also missing a "micromass challenge"

 

[hilbert2]

Tried to do "Integrate[HeavisideTheta[x^2+y^2-0.25]*HeavisideTheta[x^2+z^2-0.25]*HeavisideTheta[y^2+z^2-0.25],{x,-0.5,0.5},{y,-0.5,0.5},{z,-0.5,0.5}]" with Mathematica, but it didn't evaluate... Maybe divide the domain into small but finite cubes and take the limit as the volume of the discrete cubes approaches zero...

 

[Nidum]

Changed post content from words to pictures :

Just for interest :

Volume of one pyramid = 0.157 in³

Volume of six pyramid assembly = 0.942 in³

Volume of skeletal cube = 0.058 in³

There's almost nothing left of the cube .

 

[Charles Link]

Tried to do "Integrate[HeavisideTheta[x^2+y^2-0.25]*HeavisideTheta[x^2+z^2-0.25]*HeavisideTheta[y^2+z^2-0.25],{x,-0.5,0.5},{y,-0.5,0.5},{z,-0.5,0.5}]" with Mathematica, but it didn't evaluate... Maybe divide the domain into small but finite cubes and take the limit as the volume of the discrete cubes approaches zero...

@hilbert2 A numerical software solution is possible=I have done it=you can do 100 increments in each direction making for one million small cubes and, using 3 nested "do" loops, test each cube with inequalities to see if it is outside of the 3 cylinders. If it is, you count N=N+1. This takes only about 15 lines of code, but we are looking here for the algebraic/calculus solution=the exact answer. :) :) (Note: using enough increments in the numerical solution, I believe I have previously gotten the numerical solution to agree with the exact solution to about 8 decimal places.)

 

[Charles Link]

I am going to provide an hint on this problem that may be helpful: I think most people who look at the problem quickly figure out that the 8 corners are identical. One thing that can make it somewhat simpler is to recognize that the corners have a symmetry, e.g. about y=x. If you look at the lowest corner where x>y, (assuming the cube lies in the octant where x,y,z >0), you will see that the height z of this section is determined by the hole that heads in the y direction. (You only have one hole to consider to determine the height in the integral $\int z \, dxdy$. )

 

[MAGNIBORO]

following you hint i tried
$$\frac{V}{16} = \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}\int_{0}^{x} C_{1} \, dydx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\int_{0}^{\sqrt{x-x^2}+\frac{1}{2}} C_1 \, dydx$$

where $C_1$ is the hole, I'm going to continue the problem tomorrow >:D

 

[Charles Link]

following you hint i tried
$$\frac{V}{16} = \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}\int_{0}^{x} C_{1} \, dydx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\int_{0}^{\sqrt{x-x^2}+\frac{1}{2}} C_1 \, dydx$$

where $C_1$ is the hole, I'm going to continue the problem tomorrow >:D

So far so good. :) You should find the integrals workable=trigonometric substitutions, along with other substitutions can make them simpler. Feel free to use an integral table if need be, but evaluating the integrals with your own efforts is preferred.

 

[jedishrfu]

My guess for the volume is:

$(1)^3 - 3*\pi r^2 + 2*$(steinhold_solid_volume)

1^3 is the cubes volume
$3*\pi r^2$ is the volume of the three cylinders where r=1/2
2*steinhold volume to counteract the volume taken by two of the cylinders

Where the steinhold solid volume was computed in this post.

https://www.physicsforums.com/threads/volume-of-3-intersecting-cylinders.315248/

 

[Charles Link]

My guess for the volume is:

$(1)^3 - 3*\pi r^2 + 2*(steinhold solid volume)$

1^3 is the cubes volume
3*\pi r^2 is the volume of the three cylinders where r=1/2
2*steinhold volume too counteract the volume taken by two of the cylinders

I'm not sure what "steinholdsolidvolume is" but your guess is on the right track. If you can specify what "steinholdsolidvolume" is, you may have the right answer, but here we are looking for a calculus derivation, unless you can provide an alternative solution. :)

 

[jedishrfu]

Look at the link to an old PF thread computing the steinhold volume that I provided while you posted your response.

https://www.physicsforums.com/threads/volume-of-3-intersecting-cylinders.315248/

@benorin computed the volume integral there.

 

[Charles Link]

Look at the link to an old PF thread computing the steinhold volume that I provided while you posted your response.

https://www.physicsforums.com/threads/volume-of-3-intersecting-cylinders.315248/

@benorin computed the volume integral there.

The volume of three intersecting cylinders is a related problem. Using this result, along with the result from the intersection of two cylinders, you can solve for the remaining volume which is the form that this problem was presented. I did manage to find at least one other post that previously addressed this topic on Physics Forums. As stated in the rules, the participants are not allowed to google the answer. Others I'm sure have previously seen and/or solved this problem. The whole category of such problems, for those who wish to google it, is the "Steinmetz Solid", but googled solutions will not be credited.

 

[jedishrfu]

Sorry I didn't google it directly, I initially thought the steinmetz solid was a sphere and so checked for it and found the PF post. I then combined my answer with this piece to get the final exact solution.

Since I'm a mentor I didn't post my whole answer but wanted to see if I was on track.

I'll refrain from participating in the future.

 

[PeroK]

Error. Solution soon. Hopefully!

 

[PeroK]

I think I've fixed the error.

Spoiler

I'll do it for a cube of side length $2$.

We have three conditions for points that remain in the solid:

$x^2 + y^2 \ge 1, \ x^2 + z^2 \ge 1, \ y^2 + z^2 \ge 1$

In the first octant ($x, y, z \ge 0$) we have a symmetry. There is six times the volume of the section where $0 \le x \le y \le z$. With this restriction our integral becomes:
$$\int_{\frac{1}{\sqrt{2}}}^1 \int_{\frac{1}{\sqrt{2}}}^{z} \int_{\sqrt{1-y^2}}^{y} dx \ dy \ dz$$
Note that we must have $z, y \ge \frac{1}{\sqrt{2}}$ as these are larger than $x$.

Evaluating this integral gives $$\frac12[\frac13 (1 + \sqrt{2}) + \sin^{-1}(\frac{1}{\sqrt{2}}) - \frac{\pi}{2}]$$
The volume of one corner is $6$ times that, hence the whole solid is $48$ times that, but this solid is 8 times larger than the one in the problem, the volume of which must be:
$$1 + \sqrt{2} - \frac{3\pi}{4} \approx 0.058$$

 

[Charles Link]

If it is ok with the participants, I think I will credit a couple of correct solutions, rather than making this a race to solve a couple of integrals correctly and process all the terms that arise. Those who are actively working on this, please continue your solution, even if someone else posts a correct result.

 

[MAGNIBORO]

in post #9 i make a mistake with the upper bound of y in the second integral,
And being more clear
the $C_1$ must be the cylinder $\left ( z-\frac{1}{2}\right )^{2} + \left ( y-\frac{1}{2}\right )^{2} = \left ( \frac{1}{2} \right )^{2}$

and taking the bottom part would be $z=\frac{1}{2}-\sqrt{y-y^{2}}$
so:
$$\frac{V}{16} = \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}\int_{0}^{x} \frac{1}{2} -\sqrt{y-y^{2}} \, dydx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\int_{0}^{\frac{1}{2}-\sqrt{x-x^2}} \frac{1}{2} -\sqrt{y-y^{2}} \, dydx$$
finally
$$V=\frac{7}{3} - \sqrt{2} - \frac{\pi }{4} \approx 0.1337$$I'm not so familiar with multivariate calculus so maybe I was wrong on the bounds

 

[Charles Link]

in post #9 i make a mistake with the upper bound of y in the second integral,
And being more clear
the $C_1$ must be the cylinder $\left ( z-\frac{1}{2}\right )^{2} + \left ( y-\frac{1}{2}\right )^{2} = \left ( \frac{1}{2} \right )^{2}$

and taking the bottom part would be $z=\frac{1}{2}-\sqrt{y-y^{2}}$
so:
$$\frac{V}{16} = \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}\int_{0}^{x} \frac{1}{2} -\sqrt{y-y^{2}} \, dydx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\int_{0}^{\frac{1}{2}-\sqrt{x-x^2}} \frac{1}{2} -\sqrt{y-y^{2}} \, dydx$$
finally
$$V=\frac{7}{3} - \sqrt{2} - \frac{\pi }{4} \approx 0.1337$$I'm not so familiar with multivariate calculus so maybe I was wrong on the bounds

I missed the one sign error you had in post #9, but I will need to look over your latest input. In any case, your answer is incorrect=please check it over and try again. :) $\\$ Editing: A quick input for you is the height z ($C_1$ in your post #9) will only depend upon x and not y. (You need to replace the $y$ in the integrands with an $x$.)

 

[Charles Link]

@MAGNIBORO Please see the edited part of my post #20.

 

[MAGNIBORO]

@MAGNIBORO Please see the edited part of my post #20.

i plot the function $(z-0.5)^2 + (y-0.5)^2 = (0.5)^2$
in geogebra and look like the right function, But the camera was rotated 90 degrees...


thanks for the help,I Would have continued to make the same mistake.

 

[Charles Link]

i plot the function $(z-0.5)^2 + (y-0.5)^2 = (0.5)^2$
in geogebra and look like the right function, But the camera was rotated 90 degrees...


thanks for the help,I Would have continued to make the same mistake.

With the replacement of $y$ with an $x$ in the integrand, unless I overlooked something, everything else is completely correct that you have so far. If you evaluate it correctly, you should get the right answer. :) :)

 

[Charles Link]

On this problem which involves the intersection of three cylinders, on a second/third look at the problem, the question came to mind, (perhaps others have asked this themselves), might the volume common to the three cylinders be spherical in shape? It might be obvious to some that it isn't, but I had to entertain this possibility. Every point inside the sphere of radius $R$ will be common to the three cylinders, but are there points outside the sphere of radius $R$ that lie within the 3 cylinders? One such point is $x=y=z=(3/5)R$. $x^2+y^2+z^2>R^2$ for this point, but $x^2+y^2<R^2$ and $x^2+z^2<R^2$ and $y^2+z^2<R^2$ for this point. Thereby the volume common to all 3 cylinders is not spherical in shape. Maybe this is already obvious to many, but I thought it is worth mentioning.

 

[Charles Link]

@MAGNIBORO and anyone else who might be working at solving the integrals: There is one hurdle here with the limit in the form $(1/2)-\sqrt{2}/4$. This limit becomes much simplified if you do the right trigonometric substitution. I could give a little more detail if need be, but you might find this hint helpful. :)

 

[MAGNIBORO]

$$\frac{V}{16} = \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}\int_{0}^{x} \frac{1}{2} -\sqrt{x-x^{2}} \, dydx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\int_{0}^{\frac{1}{2}-\sqrt{x-x^2}} \frac{1}{2} -\sqrt{x-x^{2}} \, dydx$$
$$= \int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}} x \left ( \frac{1}{2} -\sqrt{x-x^{2}} \right ) \, dx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2}\left ( \frac{1}{2} -\sqrt{x-x^{2}}\right )^{2} \, dx$$
$$=\frac{3}{32}-\frac{\sqrt{2}}{16} -\int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}x^{\frac{3}{2}}\, \sqrt{1-x} \, dx + \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2} \, \frac{1}{4} -\sqrt{x-x^2} +(x-x^2) \, dx$$
$$=\frac{3}{32}+\frac{5\sqrt{2}}{16} -\int_{0}^{\frac{1}{2}-\frac{\sqrt{2}}{4}}x^{\frac{3}{2}}\, \sqrt{1-x} \, dx - \int_{\frac{1}{2}-\frac{\sqrt{2}}{4}}^{1/2} \sqrt{x-x^2} \, dx$$
$x = sin^{2}(u)$ in the 2 integrals
$$=\frac{3}{32}+\frac{5\sqrt{2}}{16} -2\int_{0}^{\frac{\pi }{8}}sin^{4}(u)\, cos^2(u) \, dx -2 \int_{\frac{\pi }{8}}^{\frac{\pi }{4}} sin^{2}(u)\, cos^2(u) \, dx$$
$$=\frac{3}{32}+\frac{5\sqrt{2}}{16} -2\int_{0}^{\frac{\pi }{8}}\left [1 - cos^{2}(u)\right ]^2\, cos^2(u) \, dx -2 \int_{\frac{\pi }{8}}^{\frac{\pi }{4}} \left [1 - cos^{2}(u)\right ]\, cos^2(u) \, dx$$

expanding and calculating the integrals with the trick $cos^2(x) = \frac{1+cos(2x) }{2}$ we get

$$\frac{V}{16} = \frac{\sqrt2+1}{16}-\frac{3\pi}{64}$$
$$V= \sqrt2+1 - \frac{3\pi}{4}\approx 0.058019$$

Very good problem, I hope there are more like these ;)

edit: the area of the intersection of the 3 cylinder is $\frac{3\pi}{4} - \sqrt2$ this look very nice

 

[Charles Link]

@MAGNIBORO Very good ! Excellent ! You got the correct answer ! A+ + $\\$ Incidentally, the total volume that gets removed is $\frac{3}{4} \pi-\sqrt{2}$. Some of this is simply due to one of the holes. There are also regions of this that are common to two of the holes. And then there is part of this that is common to all 3 holes. If I get a chance I will post what these different amounts are. (I computed them previously.) :) :) $\\$ Editing: I could use a diagram of three intersecting rings, with the region C common to all 3 rings, and Qty. 3 B type regions where two of them are intersecting, but not all 3, and Qty. 3 A type regions where the holes(rings) don't intersect with either of the other two rings: $V_{remaining}=1-3A-3B-C =1+\sqrt{2}-\frac{3}{4} \pi$ from the calculation which you just completed. A separate calculation (like the one you just completed) with just two drilled holes gives that the remaining volume is $V_{r2}=1+\frac{2}{3}-\frac{\pi}{2}$, so their volume of intersection is $V_2=\frac{2}{3}=B+C$. Meanwhile $\frac{\pi}{4}=A+2B+C$ for a single hole. A little algebra gives that the volume $C=2-\sqrt{2}$ is common to all three holes, the volume $A= \frac{\pi}{4}+\frac{2}{3}-\sqrt{2}$, and the volume $B=\frac{2}{3}-2+\sqrt{2}$. (This is also a more complete answer to the question @jedishrfu asked in post #11 and #13. The volume $C$ had been computed in another thread, but it would still be necessary to compute the volume $V_2=B+C=\frac{2}{3}$ in order to find the remaining volume from the 3 holes.)

 

[maline]

A simpler way: each corner region can be split into a cube (with an edge touching each cylinder) and three "prongs". Each prong lies along an edge of the large cube and is bounded by a face of the small cube and two of the cylinder. The cross section of the prong in the direction parallel to the small cube face is square, so the volume of the prong is one easy integral.

 

[PeroK]

A simpler way: each corner region can be split into a cube (with an edge touching each cylinder) and three "prongs". Each prong lies along an edge of the large cube and is bounded by a face of the small cube and two of the cylinder. The cross section of the prong in the direction parallel to the small cube face is square, so the volume of the prong is one easy integral.

Let's see it, then!

 

[maline]

Let's see it, then!

Sorry, my Latex is not good enough... that's why I waited till another correct solution was posted. But I think my description should make it pretty straightforward.

 

[Charles Link]

Sorry, my Latex is not good enough... that's why I waited till another correct solution was posted. But I think my description should make it pretty straightforward.

From what was described, I think what @maline is referring to is perhaps $V=8 \int_{0}^{\frac{1}{2}} x^2 \, dz$ where $(x-\frac{1}{2})^2+(z-\frac{1}{2})^2=\frac{1}{4}$. I will need to give it a try. $\\$ Editing. It's a little more complicated than that because of the hole in the z-direction. Something like that might work, but I think it's still going to be somewhat complex. $\\$ Further editing: I evaluated what I just wrote out, and it is a correct solution for the case of two holes. Perhaps some further refinement will make it work for 3 holes, but I think the integrals will not be nearly as simple.

 

[maline]

No, my solution is much simpler than that & takes all three holes into account. Should I write it out by hand and post it?

 

[Greg Bernhardt]

Should I write it out by hand and post it?

Yes. Use latex if possible here.

 

[Nidum]

@maline : Is this what you meant with your 'small cube' geometry description ?

With the corner cube gone the remaining wing shapes are all just elongated square section blocks cut away by cylindrical surfaces ?

 

[maline]

Ok, I think I can do it using a couple of diagrams.

I am assuming R=1 rather than 1/2, for simplicity. Therefore we will calculate the volume of only one of the 8 corner pieces, the one where x,y,z>0.

Consider the point x=y=z= √½. It is clear that this point is on the boundary of all 3 cylinders, so that the cubical region √½ ≤ x,y,z ≤ 1 is fully within our solid. Its volume is

(1-√½)³=1-3√½+3⋅½-(½)^3/2)=5/2-7/2⋅√½

Now we subtract the cube from our solid. Note that each of the three edges of this cube, x=y=√½, y=z=√½, & x=z=√½, lies on the boundary of one of the cylinders, parallel to its axis. So after subtracting the cube, we are left with three disconnected regions that I called "prongs", each one lying along an edge of the large cube. You can see the prongs clearly in Nidum's post #2. Since the three are identical, we will consider only one of them, the one along the edge x=y=1. It is bounded on the "outside" by the planes x=1 & y=1, and "on top" by the plane z=√½, which is a square face that it shares with the small cube. Its "inner" boundaries are the curves surfaces of the two "horizontal" cylinders. It touches the "vertical" cylinder only at the the corner point x=y=z=√½. The point of the prong extends "downward" from z=√½ to z=0 where it narrows to a point.

Here is a diagram of a cross section parallel to the xy plane, with 0 < z < √½:

https://drive.google.com/file/d/0B3BoyrPDqVLwc2FRX25fVnc0UkU/view?usp=sharing

The circle is the cross-section of the "vertical" cylinder. The lines in the x and y directions are the edges of the "horizontal" cylinders. The shaded square in the upper right-hand corner is the cross-section of our prong. It should be clear that the cross-section is indeed a square, with side 1-√(1-z²). Therefore its area is

(1-√(1-z²)²=2-z²-2√(1-z²)

To get the volume of the prong, we just integrate from z=0 to √½. The integral of "2" gives 2⋅√½=√2; the integral of "-z²" gives -⅓⋅(½)^3/2=-1/6⋅√½; and we are left with -2 times the integral of √(1-z²). But that integral can be solved geometrically:
https://drive.google.com/file/d/0B3BoyrPDqVLwb2NlMVl5MlVaQzg/view?usp=sharing

The shaded wedge is 1/8 of a unit circle, with area π/8, and the triangle is half a square of side √½, so its area is 1/4. Altogether, the volume of the prong is

√2-1/6⋅√½-2⋅(π/8+1/4)=-½+11/6⋅√½-π/4

Multiplying by 3 for three prongs, & adding the small cube:

3⋅(-½+11/6⋅√½-π/4)+5/2-7/2⋅√½=
=1+√2-¾π

 

[maline]

Oh, now that I finished my post I saw's Nidum's new beautiful diagrams. Yes, that's what I mean by "prongs". Too bad it's hard to see that their cross-section is square.
Thank you!

 

[Nidum]

The cross section of anyone wing at any point along the length is square and the function defining the edge length is a simple geometric one ?

 

[maline]

The cross section of anyone wing at any point along the length is square and the function defining the edge length is a simple geometric one ?

That's right.

 

[Nidum]

 

[Charles Link]

@maline A very good solution! I will ask @Greg Bernhardt to give you credit along with @MAGNIBORO for also having a successful solution. And thank you @Nidum for some very helpful diagrams! :) :)

 

[MAGNIBORO]

@maline if you latex is bad you can use: https://www.codecogs.com/latex/eqneditor.php?lang=en-us

@Nidum What is the program you use to make the diagrams?

 

[Nidum]

Autodesk Fusion360

 

[JasonWuzHear]

Here's an imgur album of my work:

http://imgur.com/a/AttQ7

It's not at all rigorous and tidied up, but I believe it easily leads to a solution. I'll type out what is said in my notes and post a more rigorous solution when I'm away from job. I'm too lazy to integrate right now, but I'll leave that up to future Jason and all you people out there.

I decided to integrate the area of the holes from top-down to find the volume of the holes, then double it, then subtract from the volume of the cube. To integrate the area to get a volume of the holes, I just created a coordinate, z, which will parametrize moving from top to the middle of the cube. I will integrate areas of constant z times dz to get the volume.

To better understand what these areas look like, I imagined slices of these figures at constant z. At the z=0 slice, the area is a circle. At the z=r slice, the area is a square. Any slice in between looks like a circle + some curvy triangles, as shown in the images.

Analyzing the area of the circle + the curvy triangles is easy until the triangles begin to meet and overlap. At which point, you have to subtract the overlapping portions from the area. First I will talk about the area before the triangles overlap.

The area of the curvy triangles can be found by integrating some height away from the circle, h(x), times a small piece of dx from 0 to x(z). x is shown in the images, and grows with z. For example, x(z=0) = 0, and x(z=r) = r. x(z) is just the length of the curvy triangle. There area is then the area of a circle plus 8 curvy triangles before they start overlapping.

When the two curvy triangles meet, which is at x = r/(sqrt2), they begin to overlap with each other. The extra area is in the shape of an "L" type figure, which has an area of two rectangles minus a square. The images show how simple it is to find this area and find it using the variables l(z) and s, where l(z) is the length of the square and s is the width of the rectangle. The area then will be the circle, plus the eight curvy triangles when x=r/(sqrt2), plus the 8 rectangles, minus 4 little squares.

So to find the volume, we just integrate these two areas in their proper bounds times dz from z=0 to z=r. This gives us half the volume of the holes, which we then double to find the total volume of the holes. We then subtract this volume from the volume of the cube to find the volume of what's left of the cube.

Cheers y'all, hopefully you can follow it. Good luck with your solutions!EDIT: Found some errors in my work. Will update later when I get to my desktop at home. The procedure should hold true, though.

 

[PhysicsOfLearning]

How come it is not:

1'' = 2.54cm = diameter

2.54÷2 = Radius = 1.27cm

π(1.27)² = 5.067074... = a₁

(1 side of the squares circular, drill area, which can be substituted from the area of the face of the square):

area of sq.face = 2.54² = 6.4516cm = a₂

a₂ - a₁ = 6.4516 - 5.0670 = 1.3845

Cubing the answer:

1.3845³ = 2.654cm³

The volume leftover from the drill holes in the square is:

2.654÷2.54 = 1.044 cubic inches of cube left over.

Is that correct?

 

[PeroK]

The volume leftover from the drill holes in the square is:

2.654÷2.54 = 1.044 cubic inches of cube left over.

Is that correct?

Given that the original cube only has a volume of 1 cubic inch, probably not!

Cubing an area does not give a volume.

 

[PhysicsOfLearning]

Given that the original cube only has a volume of 1 cubic inch, probably not!

Cubing an area does not give a volume.

Oops!

Methodology wrong as well?

 

[PeroK]

Oops!

Methodology wrong as well?

Yes. Look at the graphic in post #2 to see what you are dealing with.

 

[JasonWuzHear]

okay I have my answer, the volume of what's left of the cube is...0.0580191... in^3here's an imgur album with my work:
http://imgur.com/a/Xzwjk

agh forgot to say that the 4 pi r^2 should be a pi r^2 obviously. When I evaluated the integral numerically in mathematica I fixed it. Seems I got the right answer from reading the past pages, but less efficiently.

 

[Charles Link]

okay I have my answer, the volume of what's left of the cube is...0.0580191... in^3here's an imgur album with my work:
http://imgur.com/a/Xzwjk

agh forgot to say that the 4 pi r^2 should be a pi r^2 obviously. When I evaluated the integral numerically in mathematica I fixed it. Seems I got the right answer from reading the past pages, but less efficiently.

The integrals actually can be computed in closed form to give $V=1+\sqrt{2}-\frac{3}{4} \pi$. It might interest you that you can also get a numerical answer with about 15 lines of computer code using 3 nested For-Next loops ("Do" loops) that divide the volume into 100x100x100 parts and using inequalities to test whether a point (tiny cube) lies inside or outside the 3 cylinders. If it lies outside the 3 cylinders, you count $N=N+1$. Once the "Do" Loops have completed all 1,000,000 points, you compute $V=N/1,000,000$. With 1,000,000 points (100 intervals on each Do Loop), you can get an answer accurate to about 3 or 4 decimal places or more.

 

[mfb]

You can speed up the code by a factor about 4 if you remove one cylinder from the loops. You don't have to add "0" 100 times.
Dividing 1/4 of the cube surface into a 1000x1000 grid and evaluating the length of solid material for one corner for each point should give an even better approximation with a similar computing time.