# Computing volume for a scutoid

I was challenged by someone to compute the volume for a scutoid. At first I wanted to know how this shape is described mathematically, but did not find much on that so I started with what I already know:

• bottom is hexagonal
• top is pentagonal
• there is a triangle between 2 "faces"
• the 2 "faces" where the triangle cannot be in one plane

Starting from these, the first thing that I needed to do, independent of the method of computation, was to get some formula for the non-plane faces. My attempt to solve this relied on assuming that the surface satisfies Laplace equation (I thought of it as an elastic membrane that is stretched so that it touches the edges of the face I'm interested in).

Having those 2 computed, I guess one can try to integrate but I'm still traumatized by this approach, so I would make use of the method of inserting this shape into a cube like volume, generate random points and them count the ones that fall inside the shape.

Any other ideas that are not based on using a PC (solving the Laplace equations and the volume computation part)?

• There are practical ways, i.e. building one and then sinking it and measuring the volume of water displaced. Can't help beyond that. Aug 7, 2018 at 15:43
• The Wikipedia article says the name comes from scutum so I guess it's pronounced "skew-toid" not "scut-oid." Aug 7, 2018 at 15:56
• You need more constraints, there are many types of scutoids, so that information alone won't help you find the volume. Aug 7, 2018 at 15:59
• Ignore curved faces. Add a tetrahedron to get a hexagonal prism. Aug 7, 2018 at 16:00
• @EdPegg He could use the lattice points you used in your other question, but that won't have the packing property that the scutoid has because of flat faces. Aug 7, 2018 at 16:02

If you're going to treat it like a polyhedron, the volume would be the same as the volume of a frustum added to the volume of a prismatoid. This means the volume would follow the formula $$V = \frac{h_1(A_1 + \sqrt{A_1A_2} + A_2)}{3} + \frac{h_2(A_2 + 4A_3 + A_4)}{6}$$ where $$h_1$$ is the height of the frustum, $$h_2$$ is the height of the prismatoid, $$A_1$$ is the base of the frustum that isn't attached to the prismatoid, $$A_2$$ is the base they share, $$A_3$$ is the cross-sectional area of the prismatoid at half of its height, and $$A_4$$ is the base of the prismatoid that isn't attached to the frustum.