# Integration for the Volume of a Tetrahedron

In order to make a guess about the hyper volume of a hyper-tetrahedron, I was trying to find the volume of a tetrahedron using integration (it would obviously be much easier to use geometry, but it's a little tricky to extend geometry to higher dimensions).

Since I didn't really want to have to find the height of the tetrahedron, I decided to integrate up the slant altitude of one of the faces of the tetrahedron. I let $h$ be the height up the side of the tetrahedron, $l$ be the side length of the tetrahedron, and $w$ be the width of one of the the equilateral triangles at a height of $h$ from its base.

Using some easy geometry, we find out that $$w=l-\frac{2}{\sqrt 3}h$$ and so, since the height of a triangular face of the tetrahedron is $\frac{\sqrt3}{2}l$, and the area of each cross-section of the tetrahedron at $h$ units "up the face" is an equilateral triangle with side length $w$, and the area of each cross-section is $\frac{\sqrt3}{4}w^2$, the volume of the tetrahedron should be $$\int_0^{\frac{\sqrt 3}{2}l}\frac{\sqrt3}{4}w^2 dh$$ $$\int_0^{\frac{\sqrt 3}{2}l}\frac{\sqrt3}{4}(l-\frac{2}{\sqrt 3}h)^2 dh$$ But this is not correct. Where is my mistake?

I'm guessing that it has something to do with deciding to integrate up the middle of one of the faces rather than up the altitude. If so, how could I set up an integral that goes up the side of one of the faces?

• In your integral $dh$ should be replaced by $\sin\theta\, dh$, where $\theta$ is the angle between two faces. That happens because the volume of a "slice" with cross section $(\sqrt3/4)w^2$ is $(\sqrt3/4)w^2 \sin\theta \,dh$. – Aretino Jun 26 '17 at 21:23
• I suppose this approach could suffer from circular reasoning, but if you take as given that the volume of a paralletope defined by a set of vectors that define its edges is the absolute value of a particular determinant, you could then reason that the volume of a hyper-tetrahedron which shares a base and slant altitude with the paralleletope is a fixed fraction of that volume. It’s not too difficult to find that fraction inductively by computing the volumes of unit hyper-tetrahedrons. – amd Jun 26 '17 at 23:28