<s>[INST] Volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$ 
Calculate the volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$.

My approach:
For $z = 0$:
\begin{equation*}
2x + y = 4 \implies y = -2x + 4
\end{equation*}
and for $y = 0$:
\begin{equation*}
2x = 4 \implies x = 2
\end{equation*}
So the integral for the volume are:
\begin{equation*}
  \int_0^2 \int_0^{-2x +4} \left(4 - 2x - y\right)\,dy \,dx
      = \frac{16}{3}
\end{equation*}
Is my approach correct? If not, can you provide the correct answer?
 [/INST] Your approach is correct and you can calculate it also as
$$\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}\int\limits_{0}^{4-2x-y}\,dy\,dx\,dz=\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}(4-2x-y)\,dy\,dx =\int\limits_{0}^{2}\int\limits_{0}^{4-2x}(4-2x-y)\,dx\,dy$$
</s>