Tetrahedron

What is the average volume of a tetrahedron whose four vertices are chosen at random inside the unit cube? Here is a picture of one such tetrahedron.

tetrahedron

Here is the result of 1000000 trials repeated 10 times:

    \begin{equation*} \begin{array}{ccccc} 0.01384542 & 0.013850796 & 0.013834551 & 0.0138485289 & 0.013835306 \\ 0.013837212 & 0.013849963 & 0.013829154 & 0.013824425 & 0.013823698% \end{array} \end{equation*}

The volume of the tetrahedron with vertices \mathbf{x}=\left( x_{1},x_{2},x_{3}\right), \mathbf{y}=\left( y_{1},y_{2},y_{3}\right), \mathbf{z}=\left( z_{1},z_{2},z_{3}\right), \mathbf{w}=\left( w_{1},w_{2},w_{3}\right) is given by

    \begin{eqnarray*} V &=&\frac{\left\vert \left( \mathbf{x}-\mathbf{w}\right) \cdot \left( \left( \mathbf{y}-\mathbf{w}\right) \times \left( \mathbf{z}-\mathbf{w} \right) \right) \right\vert }{6} \\ &=&\frac{1}{6}\left\vert x_{1}y_{2}z_{3}-x_{1}y_{3}z_{2}-x_{2}y_{1}z_{3}+x_{2}y_{3}z_{1}\right. \\ &&+x_{3} y_{1}z_{2}-x_{3}y_{2}z_{1}-x_{1}y_{2}w_{3}+x_{1}y_{3}w_{2} \\ &&+x_{2}y_{1}w_{3}- x_{2}y_{3}w_{1}-x_{3}y_{1}w_{2}+x_{3}y_{2}w_{1} \\ &&+x_{1}z_{2}w_{3}-x_{1}z_{3} w_{2}-x_{2}z_{1}w_{3}+x_{2}z_{3}w_{1} \\ &&+x_{3}z_{1}w_{2}-x_{3}z_{2}w_{1}- y_{1}z_{2}w_{3}+y_{1}z_{3}w_{2} \\ &&+\left. y_{2}z_{1}w_{3}-y_{2}z_{3}w_{1}-y_{3}z_{1} w_{2}+y_{3}z_{2}w_{1}\right\vert \end{eqnarray*}

The average volume of the tetrahedron in the unit cube is given by the
iterated integral

    \begin{eqnarray*} &&\frac{1}{6}\int_{0}^{1}\cdots \int_{0}^{1}\,\left\vert \left( \mathbf{x}- \mathbf{w}\right) \cdot \left( \left( \mathbf{y}-\mathbf{w}\right) \times \left( \mathbf{z}-\mathbf{w}\right) \right) \right\vert \,dx_{1}\,dx_{2}\,dx_{3}\cdots dw_{1}dw_{2}dw_{3} \\ &=&\frac{3977}{216000}-\frac{\pi ^{2}}{2160}\approx 0.01 384\,277\,574 \end{eqnarray*}

That’s right, there are 12 integral signs and 12 variables.