Let $M=\{(x,y,z)\in \mathbb R^3:|x|+|y|+|z|<1\}$.
How can I determine the volume of $M$, that is, $\lambda_3(M)$?
I think it would work like this:
$$\lambda_3(M)=\int_{-1}^{1}\int_{|x|-1}^{1-|x|}\int_{|x|+|y|-1}^{1-|x|-|y|}1\ dz\ dy\ dx$$
Are there faster ways to do it?
I thought about doing the following:
Let $M_z=\{(x,y)\in \mathbb R^2:|x|+|y|<1-|z|\}$. Then
$$\lambda_3(M)=\int_{(-1,1)}\lambda_2(M_z)d\lambda(z)$$
But how can I determine $\lambda_2(M_z)$, that is the area of $M_z$?
Think of it geometrically, rather than as an integral. As @amsmath mentioned in the comments, it is a super symmetric shape. In fact, it is a regular octahedron centered at origin. We need only find the volume in the first octant. Which is the volume enclosed by the 3 co-ordinate planes and the plane $x+y+z=1$.
Hence, it's a pyramid with base a right angled triangle of area $1/2$. Hence, it's volume is ${\text{base} \times \text{height} \over 3}={1/2 \times 1 \over 3}=1/6$.
Hence, volume of $M$ is $8 \times 1/6 = 4/3$.
Edited dumb calculation thanks to @quarague.