1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Featured Challenge Math Challenge by Charles Link #1

  1. Mar 13, 2017 #1
    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.
     
    Last edited: Mar 15, 2017
  2. jcsd
  3. Mar 13, 2017 #2

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    block with holes v1.jpg

    Quite a fun shape to play with . Interesting question is whether it is now one body or eight bodies ?
     
  4. Mar 13, 2017 #3

    Charles Link

    User Avatar
    Homework Helper

    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.
     
  5. Mar 13, 2017 #4
    nice problem, I'm going to download a software to help me visualize this.
    i was also missing a "micromass challenge"
     
  6. Mar 13, 2017 #5

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    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...
     
  7. Mar 13, 2017 #6

    Nidum

    User Avatar
    Science Advisor
    Gold Member

    Changed post content from words to pictures :

    block with holes cut in half v2.jpg

    cube with hole compliment v1.jpg

    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
  8. Mar 13, 2017 #7

    Charles Link

    User Avatar
    Homework Helper

    @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.)
     
  9. Mar 13, 2017 #8

    Charles Link

    User Avatar
    Homework Helper

    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 ##. )
     
  10. Mar 13, 2017 #9
    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
     
  11. Mar 13, 2017 #10

    Charles Link

    User Avatar
    Homework Helper

    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.
     
  12. Mar 13, 2017 #11

    jedishrfu

    Staff: Mentor

  13. Mar 13, 2017 #12

    Charles Link

    User Avatar
    Homework Helper

    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. :)
     
  14. Mar 13, 2017 #13

    jedishrfu

    Staff: Mentor

  15. Mar 13, 2017 #14

    Charles Link

    User Avatar
    Homework Helper

    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.
     
  16. Mar 13, 2017 #15

    jedishrfu

    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.
     
  17. Mar 14, 2017 #16

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Error. Solution soon. Hopefully!
     
    Last edited: Mar 14, 2017
  18. Mar 14, 2017 #17

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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
  19. Mar 14, 2017 #18

    Charles Link

    User Avatar
    Homework Helper

    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.
     
  20. Mar 14, 2017 #19
    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
     
  21. Mar 14, 2017 #20

    Charles Link

    User Avatar
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Math Challenge by Charles Link #1
Loading...