Use a triple integral to find the solid of the tetrahedron T bounded by the planes x+2y+z=2, x=2y, x=0 and z=0

For z, I get z=0 and z=2-x-2y

For y, I get y=x/2 and y=1-x/2
I understand I get these from setting z=0 in the equation z=2+x+2y and using the equation x=2y and solving for y, but I am not sure why I use this method. Will someone please explain?

For x, I get the given equation of x=0
I know the second x boundary is x=1, but how do I get x=1 and why is that method used?

Projecting x + 2y + z = 2 onto the xy-plane is most easily obtained by setting z = 0.
==> x + 2y = 2.

So, T projects onto the xy-plane as the region bounded by x + 2y = 2, x = 2y, and x = 0.

Note that x + 2y = 2, x = 2y intersect as follows:
x + 2y = 2 ==> x + x = 2 (since x = 2y) ==> x = 1.

Rewrite the slanted lines as y = 1 - x/2 and y = x/2.
Finally for x in (0, 1), note that x/2 < 1 - x/2 (try x = 1/2, for instance).

Hence, the bounds of integration are
z = 0 to z = 2 - x - 2y, y = x/2 to y = 1 - x/2, and x = 0 to x = 1.

So, the volume ∫∫∫ 1 dV equals
∫(x = 0 to 1) ∫(y = x/2 to 1 - x/2) ∫(z = 0 to 2 - x - 2y) dz dy dx.

Evaluating this (just in case):
∫(x = 0 to 1) ∫(y = x/2 to 1 - x/2) (2 - x - 2y) dy dx
= ∫(x = 0 to 1) [(2 - x)y - y^2] {for y = x/2 to 1 - x/2} dx
= ∫(x = 0 to 1) [(2 - x)(1 - x/2) - (1 - x/2)^2] - [(2 - x) (x/2) - (x/2)^2] dx
= ∫(x = 0 to 1) [(1/2)(2 - x)(2 - x) - (1/4)(2 - x)^2] - [(x - x^2/2) - x^2/4] dx
= ∫(x = 0 to 1) [(1/4)(2 - x)^2 - x + 3x^2/4] dx
= (-1/12)(2 - x)^3 - x^2/2 + x^3/4 {for x = 0 to 1}
= 1/3.

I hope this helps!