Let the Tetrahedron be bounded by the planes
$x+2y+z=2$, $x=2y$, $y=0$, $z=0$
so the limits for z are easy and would be $0 \leq z \leq 2-x-2y$ I have calculated the intersection point in the $(x,y)$ plane and I get $(1, \frac{1}{2})$. So $y=0$ means that the lower boundary of $y$ should start with zero. I drew a sketch, but I am unsure about the x limits and y limits.
what would the functions $h_1$ and $h_2$ be?
$\int_{0}^{1} \int_{h_1(y)}^{h_2(y)} \int_{0}^{2-2y-x} dzdxdy$
on the other hand, if we had $x=0$ it would be easy to write because I could write it as $\int_{0}^{1} \int_{x/2}^{1-x/2} \int_{0}^{2-2y-x} dzdydx$
but since we have $y=0$ I need to determine the $h_1(x)$, $h_2(x)$, where I am confused how to do it
]1
It is easier to set up your integral as $dz \, dx \, dy$ given you have one of the equations in just $x, y$.
Here is how you can see it - if you take infinitely thin vertical strips from down to up between $y = 0$ to the plane, you realize you hit upon $x = 2y$ till $x = 1$ and then on it is the other plane $x + 2y + z = 2$. So you will have to break your integral into two.
But if you take horizontal strips, it is between two planes throughout your region finally both planes meeting at $y = \frac{1}{2}$.
$x = 2y, x + 2y + z = 2, y = 0, z = 0$
You rightly set the bounds for $z$.
Now set $z = 0$ to get the upper bound of $x, y$.
So $x = 2y, x + 2y = 2 - 2y$.
That gives you bounds of $2y \leq x \leq 2-2y$
You already know $x$ varies between $0$ and $\frac{1}{2}$ based on the intersection and your graph.
So your integral becomes -
$\displaystyle \int_{0}^{1/2} \int_{2y}^{2-2y} \int_{0}^{2-2y-x} \, dz \, dx \, dy$
You could have also set up as
$\displaystyle \int_{0}^{1} \int_{0}^{x/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx + \int_{1}^{2} \int_{0}^{(2-x)/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx$