What is the maximum volume of this object?

Question

An object has a top, side, and front view that all have a round outside border, a perfect circle of diameter = 1. What is the maximum volume this object can have?

I am talking about orthogonal views like you see in drafting.

Well, folks, I was fishing to see how many would answer with the volume of a sphere, but alas there was only one.

Answer 1

Actually, the object is not a sphere. It is a Steinmetz solid (specifically, a tricylinder), of radius 1/2. This may be seen as follows: since the front view is a circle of radius 1/2, the object must be contained within the cylinder along the x-axis of radius 1/2. Likewise, because of the top and side views, it must be contained within the cylinders of radius 1/2 along the y and z-axes. The largest solid that fulfills all of these constraints is the intersection of these three cylinders, which is a tricylinder. The volume of this object is (16 - 8√2)r³ = (16 - 8√2)/8 = 2 - √2. This is larger than the volume of the inscribed sphere.

Edit: Just for fun, here's a derivation of the volume of the solid. About 95% of the work is figuring out what integrals to use and justifying them, the integration itself takes about 5 seconds.

Derivation of volume: Consider the cube centered at the origin whose side length is 2/√2 * r. Then for a point (x, y, z) in the cube, we have |x|≤r/√2, |y|≤r/√2, and |z|≤r/√2. Then √(y² + z²) ≤ √(r²/2 + r²/2) = r, so the point is in the cylinder along the x-axis. By similar logic, it is inside the cylinder along the y-axis and the z-axis. So it is in the intersection of all three cylinders. The volume of this cube is 2√2r³, so the volume of the tricylinder is the volume of this cube plus the volume of the solid outside the cube.

Now, if (x, y, z) is outside the cube, but still in the tricylinder, then either |x|>r/√2, |y|>r/√2, or |z|>r/√2. Note that if |x|>r/√2, then since √(x²+y²) ≤ r, we necessarily have |y|r/√2} is disjoint from both {(x, y, z)∈T: |y|>r/√2} and {(x, y, z)∈T: |z|>r/√2}, and by similar logic the latter two regions are disjoint from each other. Since, by symmetry, they have the same volume, it suffices to calculate the volume of one of them and multiply it by three to find the volume of the portion of the tricylinder outside the cube. Moreover since {(x, y, z)∈T: |x|>r/√2} is the disjoint union of {(x, y, z)∈T: x>r/√2} and {(x, y, z)∈T: x<-r/√2}, and both of these regions have the same volume by symmetry, it suffices to calculate the volume of {(x, y, z)∈T: x>r/√2} and multiply it by 6.

Now, suppose (x, y, z) lies in the cross section parallel to the yz-plane at an x-coordinate greater than r/√2. Obviously since √(x²+y²)≤r, we have |y|≤√(r²-x²) and since √(x²+z²)≤r, we have |z|≤√(r²-x²). Further, as long as (y, z) is in the square of side length 2√(r²-x²), then |y|≤√(r²-x²) and |z|≤√(r²-x²), so √(y²+z²) ≤ √(2(r²-x²)) ≤ √(2(r²-r²/2) = r, so (x, y, z) is in the third cylinder as well and thus in the intersection of all three cylinders. So the cross section here is a square of length 2√(r²-x²) and thus of volume 4(r²-x²). Thus the volume of the region {(x, y, z)∈T: x>r/√2} is:

[r/√2, r]∫4(r²-x²) dx
4(r²x - x³/3)|[r/√2, r]
4(r³ - r³/3) - 4(r³/√2 - r³/(6√2))
8/3 r³ - 5√2/3 r³

Multiplying this by 6 and adding the volume of the cube yields the volume of the tricylinder:

6(8/3 r³ - 5√2/3 r³) + 2√2 r²
(16 - 8√2) r³

And we are done.

Answer 2

Good Question.

It is not a sphere. It's bigger. It surface is a set of planes bent in one direction only. Spheres bend in two directions

It is a Stienmetz solid. See http://mathworld.wolfram.com/SteinmetzSolid.html

The answer is .587

Happy New Year.

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Here is a good site for pictures of similar solids going up to an intersection of 200 cylinders. The more intersections, the more it is shaped like a sphere. http://local.wasp.uwa.edu.au/~pbourke/geometry/cylinders/

Answer 3

Couldn't it be infinity because the rest of the object could extend on forever as far as we know?

Answer 4

I think it's a sphere, so max volume = 4/3 pi r^3 where r= 1/2, so max volume = pi/6