I want to solve
$$\iiint_E 1\,dx\,dy\,dz$$
for $E=\{(x,y,z)\mid |x|+|y|+|z|\leq 1\}$.
My attempt:
I thought about it graphically. The region is like having a tetrahedron for each octant of the coordinate-system, right?
Then we would have a shape like this:
So I'd say that the integral should be $0$ because of symmetry.
But I'd like to calculate it explicitely. This is how I came up for the tetrahedron at $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$:
\begin{align*}
\int_0^1 \int_0^{1-z} \int_0^{1-y-z} 1 \,dx\, dy \,dz = 1/6
\end{align*}
But I don't know how I should change the bounds to calculate the tetrahedrons of the other areas. They should result in either $-1/6$ or $1/6$.
I'd like to have some help with that.
Best Answer
I would write the integration of the entire region, before I split into tetrahedrons. If you start with $z$, you need to integrate between $-1$ and $1$. Then based on your condition $y$ varies between $-(1-|z|)=|z|-1$ and $1-|z|$. Similarly, $x$ varies between $-(1-|z|-|y|)=|y|+|z|-1$ and $1-|z|-|y|$. Now depending on the sign, you have two regions for $x$, two for $y$ and two for $z$, total $2\cdot2\cdot2=8$ regions.
Let's choose any region, say positive $z$, negative $y$ and positive $x$. Then the smallest positive $x$ is $0$, the maximum is $1-|z|-|y|$. Since $z$ is positive and $y$ is negative, you have $$1-|z|-|y|=1-z+y$$ Then for the $y$ integration, the maximum value is $0$ and the minimum is $-(1-|z|)=z-1$. Therefore $$V_{+-+}=\int_0^1\int_{z-1}^0\int_0^{1-z+y}dxdydz$$ Following a similar reasoning, you should be able to find all the limits. Let me know if you need another example.