Find the volume of the tetrahedron bounded by the planes x+2y-z=2, x=2y, x=0, and z=0.

Question

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 $y =0$?

Answer 1

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.