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 $\left[y, \frac{-y+3}{2}\right]$ 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!
Your approach is correct but the bounds are not correct.
If you go in the order $dx, dy$, you will have to split it into two integrals. So the order I suggest is $dy, dx$.
We can rewrite $4x+2y+4z=6$ as $2x+y+2z=3$.
The volume is bound between planes $x = 0, z = 0, x = y \ $ and $ \ 2x+y+2z=3$.
If you look at the diagram, the area bound in $xy$ ($z=0$) plane, it is between lines $x = 0, y = x \ $ and $ \ 2x + y = 3$.
The bounds are $x \leq y \leq 3-2x$, $0 \leq x \leq 1$.
So the correct integral to find volume of the region is,
$\displaystyle \int_0^1 \int_x^{3-2x} \dfrac{3-2x-y}{2} \ dy \ dx$