Compute $$\iiint_Vy\,\mathrm dx\,\mathrm dy\,\mathrm dz\,,$$ where $$V=\{(x,y,z)\in\mathbb R^3\mid 2x+4y-z\le 8,x,y\ge0,z\le-4\}$$
Here is what I did and I want to know if I am right.
Let $D=\{(x, z)\in \mathbb{R}^2\mid2x-z\le8, x\ge 0, z\le -4\}$ (the projection of $V$ on the $xz$ plane). Then, $$V=\{(x, y, z)\in \mathbb{R}^3\mid (x, y)\in D, 0\le y\le \frac{1}{4}(8-2x+z)\}$$ and we have $$\iiint_Vy\,\mathrm dx\,\mathrm dy\,\mathrm dz=\iint_D\left(\int_0^{\frac14(8-2x+z)}y\,\mathrm dy\right)\mathrm dx\,\mathrm dz,$$ which is an integral that I can compute, so I won't add the calculations here.
My question is whether my set $D$, the projection of $V$ on the $xz$ plane, is correct. As far as I know, everytime I want to write such a projection I just erase the component that is $0$ on that coordinate plane and the other variables satisfy the same constraints.
$V=\{(x,y,z)\in\mathbb R^3\mid 2x+4y-z\le 8,x,y\ge0,z\le-4\}$
Plane $2x + 4y - z = 8$ intersects z-axis ($x = y = 0$) at $z = - 8$. Also at $z = -4, 2x + 4y - z \leq 8 \implies x + 2y \leq 2$.
The integral can be set up in different orders but if go in the order $dz \ dx \ dy$, the region in plane $z = -4$ is given by $ \ x, y \geq 0, x+2y \leq 2$.
For lower bound of $z$, $ \ 2x + 4y - z \leq 8 \implies z \geq 2x + 4y - 8$
So bounds are $2x + 4y - 8 \leq z \leq - 4, 0 \leq x \leq 2 - 2y, 0 \leq y \leq 1$ and the integral is,
$\displaystyle \int_0^1 \int_0^{2-2y}\int_{2x+4y-8}^{-4} y \ dz \ dx \ dy$