I am trying to find the volume of a region bounded by the following planes:

4x+2y+4z=6

y=x

x=0

z=0

I tried to first solve for z = $frac{6 – 4x – 2y}{4}$

And then I tried to set up a double integral with this function with bounds of [$y$, $frac{-y+3}{2}$] for $x$ and [0, 1] for $y$.

I got these bounds by graphing on the x-y plane the plane 4x+2y+4z = 6, since when y,z = 0, x = 3/2, and when x,z = 0, y = 3. I then found the equation for this line in the x-y plane, which is y = -2x + 3. I solved for x = $frac{-y+3}{2}$, which would be the lower bound, and the upper bound would be y, because of the plane y = x.

I then projected this intersection onto just the y-axis, where y = 1, which would be the upper bound.

I get the following integral:

$int_{0}^{1} int_{frac{-y+3}{2}}^{y} frac{6 – 4x – 2y}{4} ,dx,dy$

Is this process totally wrong? I feel like it is, but I’m not sure what I’m doing wrong. Help would be appreciated!