[bonanova]

Two circular cylinders of unit radius intersect at right angles so that their axes intersect as well.

What is the volume common to both cylinders?

What if there were three cylinders?

[Guest]

Two circular cylinders of unit radius intersect at right angles so that their axes intersect as well.

What is the volume common to both cylinders?

What if there were three cylinders?

When you say they were of "unit radius" do you mean that they both have the same radius of 1 unit?

If you do i belive the answer for the three cylinders would be

4/3pi*unit^3 because it would be bounded by a circle on all 3 planes

Im not sure about for two cylinders though

Edited July 12, 2009 by Mad Twit

[Guest]

I think it's π, found by solving the triple integral using cylindrical coordinates, so x=rcos(theta) and y=rsint(theta), boundary are 0<r<1, 0<theta<2Pi, -x^2<z<1-x^2

[dyalDragon]

a sphere, for 2 or 3 cylinders

if this is right i will come up with a proof... but math is not really my strong suit these days, which is a little silly with me being an actuary and all

[Guest]

But I'd have to say this one would be a sphere in either case, just from the way the two objects interact

[bonanova]

No one has it yet ...

Try slicing it up to be able to look inside.

[Guest]

I think the common area shape will be like that of a cylinder with r =1 and h= 2; so the volume would be

2.pi

[HoustonHokie]

I'm not sure what to call the shape of the intersecting cylinders, but it's like a regular octahedron, except that instead of straight lines from the vertex to the base, they're quarter-ellipses with one radius equal to 1 unit and the other equal to sqrt(2) units. The base is square, as would be expected in a regular octahedron, with the dimension of each side equal to 2. I did a rough integration as the minor radius goes from 0 to 1 and got the volume equals 5.3333 unit³.

[HoustonHokie]

For the 3 cylinder intersection, the shape is easiest to visualize as a cube with small pyramids on each face. The cube's dimension is sqrt(2), and the face of the cube is the base of the pyramid. Like the 2 cylinder intersection, the four connections between the base and the apex of the pyramid are not lines but quarter ellipses. This time, the radii are 1 and [2 - sqrt(2)] / 2. Another simple integration (I don't mean a real integration, but slicing it up and getting average end area volumes), and I get the volume to be 4.6863 unit³.

[bonanova]

I'm not sure what to call the shape of the intersecting cylinders, but it's like a regular octahedron, except that instead of straight lines from the vertex to the base, they're quarter-ellipses with one radius equal to 1 unit and the other equal to sqrt(2) units. The base is square, as would be expected in a regular octahedron, with the dimension of each side equal to 2. I did a rough integration as the minor radius goes from 0 to 1 and got the volume equals 5.3333 unit³.

Yup.

2/3 the volume of the [smallest] enclosing cube: edge = diameter.

2/3 x 8 = 16/3 = 5.333 ...

[bonanova]

For the 3 cylinder intersection, the shape is easiest to visualize as a cube with small pyramids on each face. The cube's dimension is sqrt(2), and the face of the cube is the base of the pyramid. Like the 2 cylinder intersection, the four connections between the base and the apex of the pyramid are not lines but quarter ellipses. This time, the radii are 1 and [2 - sqrt(2)] / 2. Another simple integration (I don't mean a real integration, but slicing it up and getting average end area volumes), and I get the volume to be 4.6863 unit³.

Right again.

Btw these animals are known as Steinmetz solids; Googling provides a look at their shapes.

8 [2 - 2^.5] r³