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Drilling 3 orthogonal holes through a cube (Part 2)

Posted on 2004-08-16
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Last Modified: 2011-04-14
I have a solid cube with its dimensions 2 units in length. I have drill-bit of radius 1 unit.
I drill a hole through the centre of each face through to the other side.
When I have finished I am left with 8 identicle corner shapes.

I would like to see the all working for the calculation of the volume of one of these corner bits?

The actual volume was provided in part 1 of the question, see :
http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_21095902.html

I would now like to see how the formula was arrived at.
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Question by:JR2003
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by:PointyEars
PointyEars earned 300 total points
ID: 11818954
I did try to calculate the volume of one of the 8 pieces left.  To visualise it, draw the three axes as follows:

                                   |  y
                                   |
                                   |*
                                   | *
                                   |   *
                                   |      *
                                   +----------*---------  x
                                 /
                               /
                             /
                           /
                         /  z

The z-axis should be at 45 degrees.  I have also drawn the hole intersection with the xy face.  Consider the radii to be == 1.  At the end, you only need to multiply the result by r³ to have the generic volume formula.

The three circles have equations:
  (x - 1)² + (y - 1)² = 1
  (x - 1)² + (z - 1)² = 1
  (y - 1)² + (z - 1)² = 1

If you draw the other two holes and imagine the three cylinders, you will see that they meet in a point that, looking at the drawing, should be a bit higher and on the right of the origin.  That point, being the intersection of the three cylinders, has all three coordinates = 1 - sqrt(2)/2.  Let's call this point C (C for cusp) and indicate with "A" the value 1 - sqrt(2)/2.

The parallel to the Z axis passing through C intersects the circle on xy in the point Pxy = (A,A,0).  Similarly, the parallel to the Y axis through C intersect the circle on xz in the point Pxz = (A,0,A).  The parallel to the Y axis through Pxy and the parallel to the Z axis through Pxz meet on the X axis in the point Px = (A,0,0).

Now, the four points C, Pxy, Pxz, and Px form a square of side A.  The volume of the pyramid formed by the square itself and the origin is given by A³/3, because the height along the X axis is precisely A.

Now we have to calculate the other funny shape left for x comprised between A and 1.  This is defined by the X axis and by three curves: an arc of circle on the xy plane, an arc of circle on the xz plane, and the arc of intersection between the two cylinders with axes perpendiclar to the xy and xz planes.

Now it is time to resolve (x - 1)² + (y - 1)² = 1.  It gives y = sqrt(2x - x²) - 1.  Similarly,
z = sqrt(2x - x²) - 1.

The intersections of the funny shape with planes parallel to the yz plane are all squares delimited by points with coordinates
  (x,0,0)
  (x,sqrt(2x - x²) - 1,0)
  (x,0,sqrt(2x - x²) - 1)
  (x,sqrt(2x - x²) - 1,sqrt(2x - x²) - 1)

The area of these intersections is therefore given by (sqrt(2x - x²) - 1)².  To calculate the volume you only need to integrate (sqrt(2x - x²) - 1)²dx for x between A and 1.  You then add A³/3 and you have the volume of 1/3 of one of the 8 pieces left after drilling the holes.

The integral should be calculable (but I am a bit rusty with those).  Whether the final result is
  8*(1 - PI*5/12)
I cannot really say.  sqrt(2x -x²) = sqrt(1 - t²) with t = x-1.  You can then possibly set t = sin(a) and get rid of the sqrt.  The PI in the result will probably come out of an arctg (or arcsin or arccos) function.

Now that you have opened a new question, I might complete the solution of the integral  :-)
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by:PointyEars
PointyEars earned 300 total points
ID: 11819836
In my previous posting I copied a wrong formula for the volume.  I said 8*(1 - PI*5/12) while the actual formula to be demonstrated is 8*[sqrt(2) - 1].  Sorry about that.

I also made another typo with bad consequences solving (x - 1)² + (y - 1)² = 1 in y.  The result is
  sqrt(2x - x²) + 1
not "- 1" as I said.

To calculate the integral of (sqrt(2x - x²) + 1)²dx for x = [A,1] we can first of all make the substitution
  t = x -1
because the result is more "readable":
  integral of [sqrt(1 - t²) + 1]²dt for t = [-sqrt(2)/2, 0]

Then we can expand the square. The resulting integral to calculate in t is
  2 - t² + 2*sqrt(1 - t²)
from -sqrt(2)/2 to 0

I saw in
  encyclopedia.thefreedictionary.com/List%20of%20integrals%20of%20irrational%20functions
that the integral of sqrt(1 - t²)dt = 1/2 * [t * sqrt(1 - t²) + arcsin(t)].  Therefore, the value of the integral is:
  1st           2nd             3rd term
  sqrt(2)  + sqrt(2)/6  +  [sqrt(2)/2 * sqrt(1 - (sqrt(2)/2)²) - arcsin(-sqrt(2)/2)]
But sqrt(1 - (sqrt(2)/2)²) = sqrt(2)/2  and arcsin(-sqrt(2)/2) = - PI/4   =>
  7*sqrt(2)/6 + 1/2 + PI/4

The volume of one of one corner should then be
  3 * [A³/3 + 7*sqrt(2)/6 + 1/2 + PI/4] =
  3 * [4/3 - 7*sqrt(2)/12 + PI/4] =
  1.29377..

This is clearly wrong because the formula should not contain PI.  Moreover, the volume of the whole cube is 8.  Therefore, each corner must have a volume significantly less than 1.

Has anybody enough patience (and perhaps 3D vision ;-) to find where I made the mistake?  Despite the addictiveness of this problem, I should take some time off from it...
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Talmash earned 200 total points
ID: 11872931
hi ,

this is realy a complicated question .

in first glance 3D regular integral will may solve .
in 2nd notice , you understand that you should choose the "dominate" diamension to integrate upon it (let say z) .

for each z , there will be a very thin surface from x to y in this shape :

***********
*          *
*     *
*   *
*  *
* *
**
*

in case of z=0 , the surface limited by x^2+y^2=1 ,
in case of z>0 you meet 2 (!!!) limitations , one in x^2+y^2=1 (created by the z axes hole)
  and there are also 2 other limitations from the other 2 holes .

this question can be solved by numeric methods .

you need an algorithm not a function / integral etc ...

tal
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by:JR2003
ID: 11876299
Thanks for all your efforts. I think this one really is a tough nut to crack.
I think this problem is a suitable candidate for a numerical method like Monte Carlo integration, testing if random points are inside or outside the volume. From the link PointyEars provided in part 1 the answer is a very simple expression so you would think it wouldn't be too difficult to solve.
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