My function is $f(x,y) = 2-|x|-|y|$ and I'm supposed to find the volume below the part of the plane which is above the $xy$-plane.
I don't understand how to find the limits of my integrals for this problem. I have tried to draw the lines for all the cases for which the absolute value of $x$ and $y$ is both positive and negative, but I don't see the limits. Is there anyone who has any suggestions?
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Hints:
For example, let us take the planes
$$\begin{cases}z=2-x-y\\z=2-x+y\end{cases}\implies y=0=\;\text{the $\;xz\,-$ plane} $$
Thus, projecting on the $\;xy\,-$ plane, we get
$$\begin{cases}y=2-x\\y=-2+x\end{cases}\implies 2x=4\implies x=2\,,\,y=0$$
Observe that for the whole four planes, you get as projection a square on the $\;xy\,-$ plane with vertices $\;(2,0),(-2,0)\;$ and etc.
Hint. The graph of this function is not a plane. However, this graph is symmetric with respect to the plane $x=0$ ($f(x,y)=f(x,-y)$) and the plane $y=0$ ($f(x,y)=f(-x,y)$). Hence it suffices to consider the case where $x\geq 0$ and $y\geq 0$ and multiply the result by $4$: $$V=4\int_{\{x\geq0, y\geq 0, x+y\leq 2\}}(2-x-y)\,dxdy =4\int_{x=0}^2\left(\int_{y=0}^{2-x}(2-x-y)\,dy\right)\,dx.$$ Can you take it from here?