$|x − y + z| + |y − z + x| + |z − x + y| = 1$, please comment on the geometrical shape

Question

If S is the outer side of the surface $|x − y + z| + |y − z + x| + |z − x + y| = 1$, please comment on the geometrical shape of the surface and volume.

I am unable to use https://www.geogebra.org/ for this particular problem.

How to find the volume of the surface by triple integral?

Answer 1

I'm only going to give you a hint. You will need to do the rest of the work, or at least show some effort of your own.

There are eight possible cases for the three absolute values in the equation. For instance, $x-y+z \ge 0$ or $x-y+z < 0$, so $|x-y+z| = x-y+z$ in the former case, and $|x-y+z| = -x+y-z$ in the latter case. The other two terms are considered the same way. So in the case where all three are nonnegative, the left-hand side simplifies to: $$(x-y+z) + (y-z+x) + (z-x+y) = x+y+z,$$ subject to the condition that $$x-y+z \ge 0, \quad y-z+x \quad 0, \quad z-x+y \ge 0$$ simultaneously. So this case describes a subset of the plane $x+y+z = 1$. It may help to determine the vertices of the region bounded by the above inequalities.

The other seven cases are handled similarly. Analyzing these gives the desired solution.

As for the volume enclosed, you will need to use other methods, but once the geometry of the region is known, not only will you know what to call the shape, you should be able to compute its volume.

Answer 2

Define

$ u = x - y + z $

$ v = y - z + x $

$ w = z - x + y $

Then in the $(u,v,w)$ coordinates, your equation becomes

$ |u| + |v| + |w| = 1 $

which is symmetric about the $vw, uv, uw$ planes. If you take $u\gt 0, v\gt0, w\gt 0 $ you get the plane

$ w = 1 - u - v $

Therefore, in the $(u,v,w)$ space, the surface is a regular octahedron.

From the defining equations of $ u,v,w $, we have

$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 && -1 && 1 \\ 1 && 1 && -1 \\ -1 && 1 && 1 \end{bmatrix}^{-1} \begin{bmatrix} u \\ v \\ w \end{bmatrix} $

Invering the matrix, gives us

$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0.5 \begin{bmatrix} 1 && 1 && 0 \\ 0 && 1 && 1 \\ 1 && 0 && 1 \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} $

Therefore, the surface in the standard basis (i.e. $(x, y, z)$ ) is an (irregular) octahedron. Its vertices can be deduce from the above relation because the vertices in $(u,v,w)$ are: $(1, 0, 0), (0, 1, 0), (-1, 0, 0), (0, -1, 0), (0, 0, 1), (0, 0, -1) $

Hence, the corresponding vertices in the our $(x,y,z)$ are

$ (0.5, 0, 0.5), (0.5, 0.5, 0), (-0.5, 0, -0.5) , (-0.5, -0.5, 0), (0, 0.5, 0.5), (0, -0.5, -0.5) $

The irregular octahedron with these vertices is plotted below

$ V = \det(A) \times 8 \times \dfrac{1}{6} (1)^3 $

where

$ A = 0.5 \begin{bmatrix} 1 && 1 && 0 \\ 0 && 1 && 1 \\ 1 && 0 && 1 \end{bmatrix} $

$\det(A) = \dfrac{1}{4} $

Hence,

$ V = \dfrac{1}{3} $