The tetrahedron has vertices $O(0,0,0); A(0,0,2); B(0,2,0); C(1,0,0)$
I was thinking that the plane $ABC$ has equation $2x+y+z=2$ since:
$\vec{BA}= \langle 0,-2,2 \rangle$ and $\vec{CA}= \langle -1,0,2 \rangle$ with a cross product $\langle 4,2,2 \rangle$
hence $4x+2y+2z=4 \cdot (1) + 2 \cdot (0) +2 \cdot (0) =4 $
So $$2x+y+z=2 \tag{1}$$
The question asked to follow the order $dz\,dy\,dx$ so I got:
$$I=\int^1_0\int^{2-2x}_0\int^{2-2x-y}_0 \; dz\,dy\,dx \tag{2}$$
Is there anything wrong with my work? I'm particularly skeptical of the $z$ limit.
Also, Is there a more straightforward way of doing it?
**edited for $dy$ limits.
Here , (1) is Correct & (2) is Partially Correct : Integral has right limits for $x$ & $z$ , though it had a wrong limit for $y$ which has since been rectified in the Question Post.
Yes , $z$ goes from minimum $0$ to maximum $2-2x-y$. We can not say that $y$ goes from minimum $0$ to maximum $2$ , it goes to maximum $2-2x$. We have no issues to see that $x$ goes from minimum $0$ to maximum $1$ here.
We have tedious integrals , though nothing much to think about now , to attack the simple triple integration :
$$I=\int_0^1\int_0^{2-2x}\int_0^{2-2x-y} 1 \; dz\,dy\,dx \tag{3}$$
$$I=\int_0^1\int_0^{2-2x} z|_0^{2-2x-y} \; dy\,dx \tag{4}$$
$$I=\int_0^1\int_0^{2-2x} (2-2x-y) \; dy\,dx \tag{5}$$
$$I=\int_0^1 (2y-2xy-y^2/2)|_0^{2-2x} \; dx \tag{6}$$
$$I=\int_0^1 (2-4x+2x^2) \; dx \tag{7}$$
$$I=(2x-4x^2/2+2x^3/3)|_0^1 \tag{8}$$
$$I=(+2/3) \tag{9}$$
Here is a visualization generated courtesy of wolfram online tool :
Alternate Way :
We can take Mirror Image of the tetrahedron to move the Corners to $(0,0,0),(1,0,2),(1,2,0),(1,0,0)$ with no change in volume.
We can see that the $YZ$ Plane has a triangle with Base $y$ & Height $y$ where $y=2x$ with $x$ varying from $0$ to $1$.
Hence the triangle Area ( $y^2/2=2x^2$ ) should be integrated to get $2x^3/3$ with $I=2/3$
Here is a visualization of the alternative way generated courtesy of wolfram online tool :
The right-most orange Area ( which is the triangle with Base $y$ & Height $y$ ) is what we want to integrate over the varying $x$ values.