# FeaturedChallenge Math Challenge by Charles Link #1

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1. Mar 13, 2017

### Greg Bernhardt

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.

Last edited: Mar 15, 2017
2. Mar 13, 2017

### Nidum

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

3. Mar 13, 2017

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.

4. Mar 13, 2017

### MAGNIBORO

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

5. Mar 13, 2017

### 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...

6. Mar 13, 2017

### 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 .

Last edited: Mar 15, 2017
7. Mar 13, 2017

@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.)

8. Mar 13, 2017

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 $\int z \, dxdy$. )

9. Mar 13, 2017

### 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

10. Mar 13, 2017

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.

11. Mar 13, 2017

### Staff: Mentor

12. Mar 13, 2017

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. :)

13. Mar 13, 2017

### Staff: Mentor

14. Mar 13, 2017

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.

15. Mar 13, 2017

### Staff: Mentor

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.

16. Mar 14, 2017

### PeroK

Error. Solution soon. Hopefully!

Last edited: Mar 14, 2017
17. Mar 14, 2017

### PeroK

I think I've fixed the error.

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:
$$1 + \sqrt{2} - \frac{3\pi}{4} \approx 0.058$$

Last edited: Mar 14, 2017
18. Mar 14, 2017

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.

19. Mar 14, 2017

### 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

20. Mar 14, 2017

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 $x$.)

Last edited: Mar 14, 2017
21. Mar 14, 2017

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

22. Mar 14, 2017

### MAGNIBORO

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.

23. Mar 14, 2017

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. :) :)

24. Mar 14, 2017

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.

Last edited: Mar 14, 2017
25. Mar 14, 2017

@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. :)