Integral over $0\leq x,y,z\leq 1$ and $x+y+z\leq 2$ What is
$$\int_{S}(x+y+z)dS,$$ where $S$ is the region $0\leq x,y,z\leq 1$ and $x+y+z\leq 2$?
We can change the region to $0\leq x,y,z\leq 1$ and $x+y+z\geq 2$, because the total of the two integrals is just
$$\int_0^1\int_0^1\int_0^1(x+y+z)dxdydz=3\int_0^1xdxdydz=\frac{3}{2}.$$
Now, can we write the new integral as
$$\int_0^1\int_{\min(2-x,1)}^1\int_{\min(2-x-y,1)}^1(x+y+z)dzdydx?$$ This gets more involved since we have to divide into cases whether $2-x-y\leq 1$ or $\geq 1$. Is there a simpler way?

You already have made two good moves: (i) replacing $S$ by the pyramidal region $S':=[0,1]^3\setminus S$, and (ii) replacing the integrand by $3x$. The integral  in question then comes to
$$Q:=\int_S (x+y+z)\>{\rm d}(x,y,z)={3\over2}-3\int_{S'}x\>{\rm d}(x,y,z)\ .$$
From a figure we read off that
$$\eqalign{\int_{S'}x\>{\rm d}(x,y,z)&=\int_0^1 x\int_{1-x}^1 \int_{2-x-y}^1 \>dz\>dy\>dx\cr &=\int_0^1 x \int_{1-x}^1 (x+y-1)\>dy\>dx\cr  &=\int_0^1 x\cdot{x^2\over2}\>dx={1\over8}\ .\cr}$$
It follows that
$$Q={3\over2}-{3\over8}={9\over8}\ .$$
