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
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Best Answer
y goes from 0 to 0.5
And for a fixed y we know that x goes from $2y$ to $2 - 2y$ because the x values are to the right of the line $x = 2y$ and to the left of the line $x = 2 - 2y$
So we end up with:
$\int_0^{0.5}\int_{2y}^{2-2y}\int_{0}^{2 - x -2y}dzdxdy$