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The volume of the region defined by the inequality : $S :$ {$ $ $(x,y,z)$ : $|x|+|y|+|z| $ $ $ $ $$\leq$$ $ $ $ $1$ $ $} is to be solved using integration . Please help .

For computer simulation , Graphing Calculator 3D Software on Windows is used , no luck on 3D plot , maybe because of the application constraints .

Nilabja Saha
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  • This is equal $8$ times the area of ${(x,\ y,\ z): x+y+z\leq1,\ x, \ y, \ z \geq 0}$. $8$ times because there's $8$ ways of choosing signs of $x,\ y,\ z$. – Jakobian May 03 '19 at 11:38
  • Try and work out what the region looks like. It would be a mistake to try and integrate it before you have figured out its shape. (Hint: it's one of the five regular polyhedra.) – TonyK May 03 '19 at 11:38
  • Did you try to sketch the region? If you consider $z=0$ you have a square in the $XY$ plane. Your region is composed of two rectangular pyramids sharing this base, with vertices at (0,0,1) and (0,0,-1). – PierreCarre May 03 '19 at 11:38
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    See https://math.stackexchange.com/questions/3063441/what-is-the-volume-of-the-region-s-x-y-z-x-y-z-%E2%89%A4-1/. – StubbornAtom May 03 '19 at 11:50

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You know x+y+z=1 is a plane with all x,y and z positive in first octant. If |x|, |y| and |z| are used, then, that plane will be trimmed to first octant only, and its mirror will be taken in all octants. So find the volume in first octant confined by this plane and the three planes formed by axes, and multiply by 8 to get the answer. For volume in first octant, we can use: $\int_0^1 \int_0^{1-z} \int_0^{1-y-z} dx dy dz$ or $\int_0^1 \int_0^{1-y} (1-x-y) dx dy$ I think the integrals evaluate to 1/6 and the answer is 4/3

Tojra
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