I know that I have to express $z$ as $z=2-x-2y$ and that's my function, but how do I find boundary points for $x$ and $y$. It says that $x$ goes from $0$ to $1$ and that $y$ goes from $0$ to $1/x/2$. Ok, $x=0$ and $y=1-x/2$ is obvious, but how do I get $x=1$ and $y =0$?
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Let $x+2y-z=2$, $x=2y$, $x=0$ and $z=0$ be an equations of planes $\pi_1$, $\pi_2$, $\pi_3$ and $\pi_4$ respectively.
Also, let $\pi_2\cap\pi_3\cap\pi_4=\{A\}$, $\pi_1\cap\pi_3\cap\pi_4=\{B\}$, $\pi_1\cap\pi_2\cap\pi_4=\{C\}$ and $\pi_1\cap\pi_2\cap\pi_3=\{D\}$.
Thus, $V_{ABCD}=\frac{S_{\Delta ABC}h}{3}$,
where $S_{\Delta ABC}$ is an area of $\Delta ABC$ and $h$ is an altitude of the tetrahedron from the vertex $D$.
Also, if know vectors then $V_{ABCD}=\frac{1}{6}\left|\vec{DA}\left(\vec{DB}\times\vec{DC}\right)\right|$
Now, just find coordinates of $A$, $B$, $C$ and $D$ and calculate the volume.
I hope it's clear now.
Michael Rozenberg
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@amd If this known then yes of course! I agree with you. But what happens if this not known? – Michael Rozenberg Oct 03 '17 at 06:24
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Umm... do exactly what you’ve outlined in the first half of this answer, of course: compute them. – amd Oct 03 '17 at 06:32