The geometry is difficult to set up, and there are many approaches. I set one of the faces to be the base. On that face, I’ll let \(b\) be the distance from a vertex of that face to the centre of the same face. I can split this face into three isoceles trianges, each of which has side lengths \(b\text{,}\) \(b\) and \(a\text{,}\) with angles \(\pi/6\text{,}\) \(\pi/6\) and \(2\pi/3\text{.}\) I apply the sine law to this triangle.
\begin{equation*}
\frac{a}{\sin \frac{2\pi}{3}} = \frac{b}{\sin \frac{\pi}{6}}
\implies \frac{a}{\frac{\sqrt{3}}{2}} = \frac{1}{2}
\end{equation*}
This implies \(b = \frac{a}{\sqrt{3}}\text{.}\) Then there is a right triangle whose vertical are one vertex of the base, the centre of the base, and the top vertex. This triangle has hypotenuese \(a\text{,}\) base \(b\) and height \(c\) where \(c\) is the height of tetrahedron, from the center of the base to the top vertex. Pythagorus allows me to calculate c: \(c = \sqrt{a^2 - b^2} = \sqrt{\frac{2}{3}}
a\text{.}\)
I’m going to slice the tetrahedron with horizontal plane slices, stacked vertical. The previous calculation for \(c\) gives us a bound for integration, so the integral goves from \(0\) to \(\sqrt{\frac{2}{3}}a\text{.}\) I’ll work with the variable \(h\) for height. Because it is convenient, I’ll say that \(h\) starts at zero at the top vertex, and goes to \(\sqrt{\frac{2}{3}}a\) at the base.
I have to do some geometry to find the area of each slice. slice. Each slice is an equilateral triangle, but what is the side length (depending on \(h\))? Recall the base triangle: it has side length \(a\) when \(h =
\sqrt{\frac{2}{3}}a\text{.}\) This side length increases linearly from \(0\) when \(h=0\) to \(a\) when \(h =
\sqrt{\frac{2}{3}}a\text{.}\) The final result gives the solve of the linear relationship. Therefore, for \(h \in [0,
\sqrt{\frac{2}{3}}a ]\text{,}\) the side length of the triangle at height \(h\) is \(\sqrt{\frac{3}{2}}h\text{.}\)
An equilateral triangle with side length \(s\) is \(\frac{\sqrt{3}a^2}{4}\text{.}\) This gives the area of each slice.
\begin{equation*}
A = \frac{\sqrt{3}}{4} \left( \sqrt{ \frac{3}{2}} h
\right)^2 = \frac{\sqrt{3}}{4} \frac{3}{2} h^2 =
\frac{3\sqrt{3}h^2}{8}
\end{equation*}
Then I can setup the integral of the slices.
\begin{equation*}
V = \int_0^{\sqrt{\frac{3}{2}}a} A(h) dh =
\int_0^{\sqrt{\frac{3}{2}}a} \frac{3\sqrt{3}h^2}{8} dh
\end{equation*}
This is a reasonable polynomial integral.
\begin{equation*}
V = \frac{3\sqrt{3}}{8} \int_0^{\sqrt{\frac{3}{2}}a} h^2 dh
= \frac{3\sqrt{3}}{8} \frac{h^3}{3}
\Bigg|_0^{\sqrt{\frac{3}{2}}a}
= \frac{\sqrt{3}}{8} \left( \sqrt{\frac{2}{3}} a \right)^3
= \frac{\sqrt{3}}{8} \frac{2\sqrt{2}}{3\sqrt{3}} a^3 =
\frac{a^3}{6\sqrt{2}}
\end{equation*}
