What is the volume of the region $S =\{(x, y, z) : |x| + |y| + |z| ≤ 1\}$?

What is the volume of the region $S =\{(x, y, z) : |x| + |y| + |z| ≤ 1\}$ ?

How can i find the volume

any hints /solution

The region $S$ looks somewhat like this:

If you consider only the first octant, you have the simplex:

Due to symmetry, the volume of $S$ is $8$ times the volume of this simplex.

That is, $\operatorname{vol}(S)=8\times $$\iiint_{\substack{x+y+z\le 1\\ x,\,y,\,z\,\ge\, 0}}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$

First of all, you have to understand what $|x|+|y|+|z|=1$ means.

Obviously this equation is fully symmetric when changing the octants. Thus one can restrict to the first one and we get then $x+y+z=1$ instead. But this is nothing but the plane through the points $X=(1,0,0)$, $Y=(0,1,0)$, and $Z=(0,0,1)$.

Therefore your set $S$ is nothing but the octahedron, centered at the origin $O=(0,0,0)$, with circumradius $$r=\overline{OX}=\overline{OY}=\overline{OZ}=1$$ i.e. with edge length $$s=\overline{XY}=\overline{XZ}=\overline{YZ}=\sqrt2$$

 
 
Next one calculates the area $A_O$ of the triangle $OXY$ as $$A_O=\frac12\ \overline{OX}\ \overline{OY}=\frac12$$ Then the volume $V$ of the octahedral octant $OXYZ$ would be $$V=\frac13\ A_O\ \overline{OZ}=\frac16$$ Thus the searched for volume of the full octahedron is $$vol(S)=8\ V=\frac86=\frac43$$

 
 
You even could derive the total surface area of that octahedron. Let $M$ be the center of $X$ and $Y$. Thus $M=(\frac12, \frac12, 0)$. Then you'll have the distances $$\overline{XM}=\overline{MY}=\frac12\ \overline{XY}=\frac{\sqrt2}2=\frac1{\sqrt2}$$ and $$\overline{ZM}^2=\overline{XZ}^2-\overline{XM}^2=2-\frac12=\frac32$$ Then the area $A_Z$ of the triangle $XYZ$ becomes $$A_Z=\frac12\ \overline{XZ}\ \overline{ZM}=\overline{XM}\ \overline{ZM}=\frac1{\sqrt2}\cdot \frac{\sqrt3}{\sqrt2}=\frac{\sqrt3}2$$ and therefore finally $$\textit{surf}\,(S)=8\ A_Z=8\cdot\frac{\sqrt3}2=4\sqrt3$$

--- rk