Volume of the solid where it is enclosed from $2x + y + z = 4$ and the planes $x = 0$, $y = 0$, $z = 0$

Question

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?

Answer 1

Your approach is correct and you can calculate it also as $$\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}\int\limits_{0}^{4-2x-y}\,dy\,dx\,dz=\int\limits_{0}^{4}\int\limits_{0}^{2-\frac{y}{2}}(4-2x-y)\,dy\,dx =\int\limits_{0}^{2}\int\limits_{0}^{4-2x}(4-2x-y)\,dx\,dy$$