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Waterman
10-25-2005, 10:37 PM
OK - I'm having problems visualizing this and solving this problem, which on the surface, seems like it should be easy.

I have a cube with length of its side equal to S. At the exact center of any face of the cube I drill (cut, burn, whatever) a perfect cylinder through the cube that has a diameter equal to S. At this point the remaining volume is:

S^3 - 1/4*(Pi)*S^3 which, after using 3.14159 for (Pi) leaves:

0.21460 S^3

Now if I go to any of the four adjacent faces and drill a second hole with diameter equal to S what is left? I am visually that the common space occupied by the two holes is a sphere of diameter S but I'm not sure how to calculate what remains of the rest of the cube.

Any help out there?

panamajack
10-25-2005, 11:06 PM
It should just be

Vcube - ( 2 * Vcylinder - Vsphere )

This might simplify to something nicer looking, I don't know.

MikeS
10-25-2005, 11:19 PM
It should just be

Vcube - ( 2 * Vcylinder - Vsphere )

This might simplify to something nicer looking, I don't know.
You seem to have used the postulate that the intersection volume of the two spheres is a sphere, but that's not true. After all, if your cylinder axes are the x-axis and the y-axis, the object is still going to have a square cross-section along the z-axis.

As to how to actually solve it, it should be a straightforward volume integral, and the only trick is setting up the limits. I'll think about it a little more before bed and see if I come up with anything.

Ringo
10-25-2005, 11:23 PM
For surface area:

Surface area of the cube prior to hole drilling = 6S²

Area drilled away with 1st drilling = 2ΠS²

Area drilled away with 2nd drilling = 2ΠS²

Area drilled away with both drillings = 2*(2ΠS²)

Surface of original cube remaining after both drillings = (6S²)-(2*(2ΠS²))

But I suspect you're looking for the remaining volume. The volume removed by your drilling is not a sphere. You need to subtract from the volume of the original cube both the cylindrical volume removed by the original (1st) drilling as well as the volume removed by the secondary drilling, which is a volume equal to the volume of the first drilled cylindar minus the volume of the intersection of those two cylinders. I've drawn it, and it's not a sphere; there are linear edges.

Calling ultrafilter.

Heh - and I'm a licensed scientist.

samclem
10-25-2005, 11:53 PM
Waterman. A question. Is this homework? I read some of your posts and you seem to be a college grad.

If this is a specific homework problem, I'm required to shut it down. :)

Just asking......

samclem GQ mod

ultrafilter
10-26-2005, 12:02 AM
For simplicity's sake, let's say that the two cylinders are given by the equations x2 + y2 < S/2 and x2 + z2 < S/2. Their intersection O is the set of points where both conditions hold. Therefore, for any point in the intersection, you have 2x2 + y2 + z2 < S.

The total volume of the remaining shape is (1 - p/2)S3 + òOdV. Setting up limits on multiple integrals was never my strong suit, so I'll leave that for someone who can do it more quickly.

Waterman
10-26-2005, 12:56 AM
Waterman. A question. Is this homework? I read some of your posts and you seem to be a college grad.

If this is a specific homework problem, I'm required to shut it down. :)

Just asking......

samclem GQ mod
Well I am quite flattered that you think I am a college grad student. I am a 57 year old working water and wastewater engineer who hasn't seen any homework in 35 years.

But thanks for asking!!! ;) :D

Waterman
10-26-2005, 01:02 AM
I got lost in conveying my 3 dimensional problem after only acting in two dimensions. I should have also expressed the desire to drill the third cylinder through the z-axis, but I am more unclear as to what is left with respect to the remaining volume of the original cube.

Now I believe, the intersection of the three cylinders is a sphere with volume of S^3?

Waterman
10-26-2005, 01:13 AM
For simplicity's sake, let's say that the two cylinders are given by the equations x2 + y2 < S/2 and x2 + z2 < S/2. Their intersection O is the set of points where both conditions hold. Therefore, for any point in the intersection, you have 2x2 + y2 + z2 < S.

The total volume of the remaining shape is (1 - p/2)S3 + òOdV. Setting up limits on multiple integrals was never my strong suit, so I'll leave that for someone who can do it more quickly.
As my response to the moderator should perhaps hint at, it has been a long time since I have done any multiple integrals.

What I see remaining are these curvilinear pieces in each of the eight vertices of the cube with the length from the vertex to the edge equal to (basically the diagonal of the face less the length of the face divided by 2):

(s*((2^0.5)-1)/2)

nivlac
10-26-2005, 01:56 AM
The volume of the intersection of two perpendicular cylinders of radius r is 16r3/3. Some beautiful pictures here (http://astronomy.swin.edu.au/~pbourke/geometry/cylinders) if you want to see what this intersection looks like.
So the answer to the OP is VOLcube - 2*VOLcyl + VOLintersection = s3(1 - pi/2 + 2/3) = .095870339s3.

Waterman
10-26-2005, 03:38 AM
The volume of the intersection of two perpendicular cylinders of radius r is 16r3/3. Some beautiful pictures here (http://astronomy.swin.edu.au/~pbourke/geometry/cylinders) if you want to see what this intersection looks like.
So the answer to the OP is VOLcube - 2*VOLcyl + VOLintersection = s3(1 - pi/2 + 2/3) = .095870339s3.
The link you provided is linked to the problem of the intersection of three cylinders and answers at least one of my original questions which is that the intersection is not a sphere of diameter equal to S

Intersection of Three Cylinders (http://astronomy.swin.edu.au/~pbourke/geometry/planelev/)

According to the above link the volume of three intersecting cylinders is 0.58579*S^3 and the volume of two intersecting cylinders (from your original link) is 0.66667*S^3. There are three intersections of two cylinders and one intersection of three cylinders therefore remaining volume after drilling the three cylinders is:

Volume of cube=S^3= 1.00000*S^3
Less volumes of three cylinders drilled out=3*(3/4)*(Pi)*S^3=2.35619*S^3
Plus volume of three intersecting pairs of cylinders=3*(2/3)*S^3=2.0000*S^3
Less Volume of intersection of three intersecting cylinders=(2-(2^0.5))*S^3=0.58579*S^3

Which equates to:
(1.0000-2.35619+2.0000-0.58579)*S^3 = 0.05802*S^3 of the original volume of the cube!

This seems to work!!

Mathochist
10-26-2005, 10:49 AM
OK - I'm having problems visualizing this and solving this problem, which on the surface, seems like it should be easy.

I have a cube with length of its side equal to S. At the exact center of any face of the cube I drill (cut, burn, whatever) a perfect cylinder through the cube that has a diameter equal to S. At this point the remaining volume is:

S^3 - 1/4*(Pi)*S^3 which, after using 3.14159 for (Pi) leaves:

0.21460 S^3

Now if I go to any of the four adjacent faces and drill a second hole with diameter equal to S what is left? I am visually that the common space occupied by the two holes is a sphere of diameter S but I'm not sure how to calculate what remains of the rest of the cube.

Any help out there?

Like other posters have said, the intersection is a sphere. However, the volume of the intersection is a pretty standard calc 3 homework example, so I'll advise you to ask a calc 3 student.

spinky
10-26-2005, 12:47 PM
Like other posters have said, the intersection is a sphere.Like other posters have said, the intersection is not a sphere. See the links provided.

spinky
10-26-2005, 12:50 PM
Also, I noticed that nivlac's link has one shape formed by cylinders intersecting the sides of things called "Waterman Polyhedra", which may be of special interest to our OP. :D

Waterman
10-26-2005, 12:57 PM
The volume of the intersection of two perpendicular cylinders of radius r is 16r3/3. Some beautiful pictures here (http://astronomy.swin.edu.au/~pbourke/geometry/cylinders) if you want to see what this intersection looks like.
So the answer to the OP is VOLcube - 2*VOLcyl + VOLintersection = s3(1 - pi/2 + 2/3) = .095870339s3.
Your link was exactly what I needed to see both the two cylinder and three cylinder intersections. It gave me the visualization that I hope resulted in my correct solution described above.

THANKS AGAIN

Mathochist
10-26-2005, 06:33 PM
Like other posters have said, the intersection is not a sphere. See the links provided.

You're right, I accidentally edited out "not". Mea culpa.

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