Finding $\iiint 6z\,dx\,dy\,dz$ over $\lbrace (x,y,z) \in \mathbb{R}^3 : |x+y| \le z \le |x|+|y|\le 1 \rbrace$

Question

I want to calculate integral : $\iiint 6z\,dx\,dy\,dz$. The area is $$\Omega= \lbrace (x,y,z) \in \mathbb{R}^3 : |x+y| \le z \le |x|+|y|\le 1 \rbrace$$

My problem is this, that I don't know what is the area and what are ranges of variable $x,y,z$.

Answer 1

Draw a figure, at least of the $(x,y)$-plane!

The given domain $\Omega$ projects to the square $Q:=\bigl\{(x,y)\bigm||x|+|y|\leq1\bigr\}$ having vertices at the points $\pm1$ on the axes. A priori this square is the "base" of the integration. But we can simplify matters: Divide $Q$ by the axes and the lines $y=\pm x$ into eight triangles. When $(x,y)$ is in the first or in the third quadrant then $|x+y|=|x|+|y|$. The condition $|x+y|\leq z\leq|x|+|y|$ then does not produce a $z$-interval of positive length, so that the triangles in these quadrants can be neglected. Furthermore all four triangles in the two other quadrants give the same value by symmetry, so that we may restrict to the triangle $\triangle:=\bigl((0,0),(0,1),(-{1\over2},{1\over2})\bigr)$. Within $\triangle$ we have $|x+y|=x+y$ and $|x|+|y|=y-x$, so that the condition on $z$ reads as $x+y\leq z\leq y-x$. In this way we obtain $$J:=\int_\Omega 6z\>{\rm d}(x,y,z)=4\int_\triangle\int_{x+y}^{y-x}6z\>dz\>{\rm d}(x,y)\ .$$ The inner integral evaluates to $$3z^2\biggr|_{x+y}^{y-x}=-12xy\ ,$$ so that we get $$J=-48\int_\triangle xy\>{\rm d}(x,y)=-48\int_{-1/2}^0\int_{-x}^{1+x}xy\>dy\>dx=\ldots=1\ ,$$ where I have used Mathematica for the last double integral.