Given $$2x + y + z = 4 \ \ x = 0,\ y = 0,\ z = 0$$ This is a plane in the first quadrant so I need to set up the triple integral and solve in order to get the area but I am not sure how to determine the bounds of each integral. This is what I have:
$$ \int_0^1\int_0^1\int_0^2 2x + y + z \ dx \ dy \ dz$$
Hints: substitute in the plane's equation to find its intersection points with the coordinate planes
$$\begin{cases}y=z=0\implies2x=4\implies x=2\\{}\\x=z=0\implies y=4\\{}\\x=y=0\implies z=4\end{cases}$$
Thus, on the $\;xy\,-$ plane we have $\;0\le x\le 2\;$ while $\;2x+y=4\implies y=-2x+4\;$ ...