The problem requires me to find the volume of the region in the first octant bounded by the coordinate planes and the planes $x+z=1$, $y+2z=2$, and here is my setup:
$$ \begin{aligned} &\phantom{\implies}x+z=1\\ &\implies z=1-x\\ \end{aligned} $$
$$ \begin{aligned} &\phantom{\implies}y+2z=2\\ &\implies y=2-2z=2-2(1-x)=2x\\ \end{aligned} $$
But the solution in the notes is as follows:
$$ \begin{aligned} V&=\int_0^1\int_0^{1-x}\int_0^{2-2x}dydzdx\\ &=\int_0^1\int_0^{1-x}\left[y\right]_0^{2-2x}dzdx\\ &=\int_0^1\int_0^{1-x}(2-2x)-(0)dzdx\\ &=\int_0^1\int_0^{1-x}2-2xdzdx\\ &=\int_0^1(2-2x)\left[z\right]_0^{1-x}dx\\ &=\int_0^1(2-2x)[(1-x)-(0)]dx\\ &=\int_0^1(2-2x)(1-x)dx\\ &=\int_0^12x^2-4x+2dx\\ &=\left[\frac{2}{3}x^3-\frac{4}{2}x^2+2x\right]_0^1\\ &=\left[\frac{2}{3}(1)^3-\frac{4}{2}(1)^2+2(1)\right]-\left[\frac{2}{3}(0)^3-\frac{4}{2}(0)^2+2(0)\right]\\ &=\frac{2}{3}\\ \end{aligned} $$
So I wonder where the upper limit of $y$, which is $2-2x$, comes from.
The region is given by the inequalities $x,y,z\geq 0$, $x+z\leq 1,y+2z\leq 2$. It is pyramid with a rectangular base of sides $1$ and $2$, which lays in the $xy$-plane, and of height $1$ because the apex is at $(0,0,1)$. By elementary geometry the volume of this pyramid is $V=\frac{B\cdot h}{3}=\frac{(1\cdot 2)\cdot 1}{3}=\frac{2}{3}$.
Such volume can be evaluated also as an iterated integral in at least two different ways:
The slices are perpendicular to the $z$-axis: $z\in [0,1]$, $(x,y)\in [0,1-z]\times [0,2-2z]$, $$\begin{aligned} V&=\int_{z=0}^1\int_{x=0}^{1-z}\int_{y=0}^{2-2z}dydxdz\\ &=\int_{0}^1(1-z)(2-2z)\,dz=2\int_{z=0}^1(1-z)^2\,dz\\ &=2\int_{t=0}^1t^2 dt=2\Big[\frac{t^2}{3}\Big]_0^1=\frac{2}{3}. \end{aligned} $$
The slices are perpendicular to the $x$-axis: $x\in [0,1]$, $z\in [0,1-x]$, and $y\in [0,2-2z]$, $$\begin{aligned} V&=\int_{x=0}^1\int_{z=0}^{1-x}\int_{y=0}^{2-2z}dydzdx\\ &=\int_{x=0}^1\int_{z=0}^{1-x}(2-2z)dzdx\\ &=\int_{x=0}^1(2(1-x)-(1-x)^2)dx\\ &=\int_{t=0}^1(2t-t^2)dt=\Big[t^2-\frac{t^2}{3}\Big]_0^1=\frac{2}{3}. \end{aligned} $$