I want to evaluate the integral $\int_T x^2 dV$, where $T$ is the tetrahedron bounded by the planes $x=0, y=0, z=0$ and $4x+2y+z=4$. My problem is with the setup with the region of integration. Would the following be correct?
$\int_0^4\int_0^{\frac{4-y}{2}}\int_0^{4-y-2z}x^2dxdydz$
Basically what I have done is rewritten for $x$ in terms of the other variables in the inner integral, then done the same with $x=0$ in the second integral, and then the same with $x=y=0$ in the last one.
Yes, $z$ can take any values in $[0,4]$. For each such $z$, since $4x+2y+z\leqslant4$, $y$ can take any value from $0$ to $2-\frac z2$. And, for any such $z$ and any such $y$, $x$ can take any value from $0$ to $1-\frac y2-\frac z4$. So, it should be$$\int_0^4\int_0^{2-z/2}\int_0^{1-y/2-z/4}x^2\,\mathrm dx\,\mathrm dy\,\mathrm dz=\frac2{15}.$$