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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 $M_z$?

user700332
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    This thing is super symmetric. Hence, you can drop all the absolute values and multiply by 8. – amsmath Sep 03 '19 at 14:23
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    Just use the volume formula to compute triangular pyramid of base area $1/2$ and height $1$, which is $1/6$. Then multiply it by 8. – mastrok Sep 03 '19 at 14:28
  • $M_z$ is a square with side length $\sqrt 2(1-|z|)$. – amsmath Sep 03 '19 at 14:34
  • @amsmath How do you see that it is a square with the side length you mentioned? – user700332 Sep 03 '19 at 14:50
  • @user700332 $M_z$ is also symmetric. So, let's consider it only for positive values of $x$ and $y$. Then ${(x,y) : y < (1-|z|)-x}$ is bounded by the coordinate axes and the line $y = (1-|z|)-x$. So, it's a triangle. Now, fuse your 4 triangles to get the square. – amsmath Sep 03 '19 at 14:58
  • Previously asked: 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 Sep 03 '19 at 16:14

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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 $x+y+z=1$. graph 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.

MathIsNice1729
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    Your computation is correct but unnecessarily complicated. You used the equilateral triangle as a base, then you need to explain how you got the area from the sides and you need to compute the height. Better to take a triangle in the xy-plane as base, that is a rectangular triangle and you can just read of the height of the pyramid. – quarague Sep 03 '19 at 15:04
  • @quarague Yeah. I'm dumb lol. Edited. – MathIsNice1729 Sep 03 '19 at 15:06