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You are currently reading a thread in /sci/ - Science & Math

You are currently reading a thread in /sci/ - Science & Math

Thread images: 5

Hello /sci/

I've been given a little thought experiment that I'd like to pass on to you, it goes like this:

You have a cube with sidelengths 1m.

You then take a drill which has a diameter of 1m, and drill through the cube from all sides, so that you end up with pic related.

Now, how do you determine the volume that is left after drilling?

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just integrate one of the points and multiply by 8

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>>7771417

>Your Calc I homework

>Calling it a thought experiment

So this is how it feels to talk with a retarded person.

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>>7771421

Just give him shit he won't understand

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8 times the small part constructed by three planes and three cylinder walls. Easy peasy.

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Use symmetry, there's 24 parts with equal shape.

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>>7771417

I heart SW

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So what's the value?

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>>7771417

archimedes had a pretty good method for finding volume

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>>7771591

I think the value may be about 0.058014 m^3, but I can't garantuee it.

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>>7771417

Figure out one of the resulting shapes, multiply by 8.

Don't know how to do that though. I'm sure the problem is probably able to be subdivided better than that.

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>>7771417

>draw it in CAD

>punch 3 holes through it

>look at model volume

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>>7771417

This is clearly your Calc I homework.

A hint: You can do this with a double integral. How would you describe the cross-section of the leftover pieces? What function determines how wide the base is?

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>>7771630

http://pastebin.com/1zHpLeVu

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>>7771417

Your pic related makes no sense. A drill would make cylinder shaped cavities in the cube, but your pic looks like a sphere has been carved out of the cube.

That's dumb.

Also, if the cube side lengths are 1m and the drill bit diameter is 1m, you're going to end up with 8 pieces of cube scraps, not a cube with holes in it.

Furthermore, read the sticky. There is now actually a homework board. But since you are probably too lazy to check, I'll provide a link.

>>>/hm/

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>>7771694

>your pic looks like a sphere has been carved out of the cube.

are you retarded, blind or both?

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>>7771694

>pic looks like a sphere has been carved out of the cube

If it had been a sphere then there would be no sharp inward pointing tips on the triangular parts. Since there are your hypothesis implodes.

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>>7771417

Imaging x going from 0 to 1/2. If y<x, then the only cylinder that is relevant is the x-z cylinder.

Break it up into 2 parts.

[math]16 \int_0^{\sqrt{2}/4} x \sqrt{1/4-x^2}\,dx \,+\, 16 \int_{\sqrt{2}/4}^{1/2} (1/4-x^2)\,dx[/math]

I get .585786, but that's hard to believe based on the picture. More than half left?

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>>7771784

>If y<x, then the only cylinder that is relevant is the x-z cylinder.

No. The perpendicular cylinders make an impact much earlier than that. In fact the perpendicular cylinders would be tangential to the face that the xz cylinder penetrates.

>I get .585786, but that's hard to believe based on the picture.

I agree. It must be far less than 1/2.

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Put the object in water

boom

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>>7771794

Indeed, I think it's more like...

[math]8 \int_0^{1/2} \int_0^{1/2-\sqrt{x(1-x)}} \min\Big(1/2-\sqrt{x(1-x)},1/2-\sqrt{y(1-y)}\Big)\,dy\,dx[/math]

Split that up, and I think you get 1 - .585786... What I had before was the volume of the removed bit.

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>>7771417

What is the volume of the cylinders? What is the shape created by the intersection of all three cylinders?

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>>7771784

>then the only cylinder that is relevant is the x-z cylinder.

wat? Why?

That's not true.

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>>7771417

Evaluate triple integral of one of the corner pieces, multiply by 8.

If you have difficulty setting up the limits, project the 3d object onto a 2d plane for each side, as if you are looking at it from above(while the flat surfaces are touching the 2d plane),

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>>7771863

>intersection

No. It has to be the union of the cylinders which indeed is what you have in your figure. And that is not an intersection.

I would however like to see what the actual union would look like.

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>>7771863

http://mathworld.wolfram.com/SteinmetzSolid.html

This is the way. Looking at the 8 corner pieces is the wrong way.

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>>7771861

>I think you get 1 - .585786

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>>7771889

I was referring to the shape created by the intersecting cylinders, so I fail to see how that is incorrect for the purposes of finding a volume. The intersecting edges of the cylinders create a definite shape, which has a definite volume, so what's the deal? Sure union may get the point across better, but there is plenty of ways to explain it. Here's another: the solid common to three right circular cylinders of equal radii intersecting at right angles.

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>>7771889

Also, this is what it looks like.

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>>7771891

I don't know what this is, but I like it either way.

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>>7771895

The volume of the cube left over is

1 - (3*(cylinder volume) - 2*(SteinmetzSolid))

Picture it as 1 full cylinder and two broken cylinders, where the solid piece missing from each broken cylinder is equal to the union of all three (steinmetz solid). Remove this volume from the cube volume and you have whats left over. It should be around .557

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>>7771894

>I was referring to the shape created by the intersecting cylinders

This I agree with. That is somewhat different from the intersection of the cylinders. I am normally not this pedantic but in this case there would be a noticeable difference.

>>7771895

>Also, this is what it looks like.

Thanks! Intriguing shape. I had tried to visualise it in my mind but failed. Seeing it now however I see it is easy to see that the shape is correct. Hindsight is so much clearer. Again.

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>>7771417

its easy, the negative space in the picture is the intersection of 3 cylinders.

To visualize the calculations, first, think about the intersection of two cylinders. If you look at the cross sections of two cylinders from above, you will see that the slices are squares (because the cross sections of a single cylinder from above are rectangles). The largest of these squares is 1 meter by 1 meter, which is well outside the perimeter of a circle with diameter 1 meter. To do the integration of the three cylinder intersection, just integrate the squares from the two cylinder intersection, but cut off some of the area of each square that expands out of the circle with diameter 1 meter.

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>>7771967

never mind, it is the union of parts of three cylinders. You can calculate this by adding the volume of the first cylinder to the second cylinder, then by subtracting the volume of the intersection of two cylinders. When you have the volume of the union of two cylinders, add it to the volume of 1 cylinder, then subtract the intersection of three cylinders.

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>>7771417

How would that even work? It wouldn't stay together if you drilled 1 meter.

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