Compute the mass $M$:
\[ M = \int_{y=0}^{1} \int_{x=0}^{2(1-y)} \rho\, dx\, dy = \int_0^1 2(1-y)\, dy = 2\int_0^1 (1-y)\, dy = 2\Big[y - \tfrac{y^2}{2}\Big]_0^1 = 2(1-\tfrac{1}{2}) = 1. \]Compute $\bar{x}$:
\[ \bar{x} = \frac{1}{M} \int_0^1 \int_0^{2(1-y)} x\, dx\, dy = \frac{1}{M} \int_0^1 \Big[\tfrac{x^2}{2}\Big]_0^{2(1-y)} dy = \int_0^1 \frac{(2(1-y))^2}{2}\, dy = 2\int_0^1 (1-y)^2\, dy. \] \[ \bar{x} = 2\int_0^1 (1 - 2y + y^2)\, dy = 2\Big[y - y^2 + \tfrac{y^3}{3}\Big]_0^1 = 2(1 - 1 + \tfrac{1}{3}) = \tfrac{2}{3}. \]Compute $\bar{y}$:
\[ \bar{y} = \frac{1}{M} \int_0^1 \int_0^{2(1-y)} y\, dx\, dy = \frac{1}{M} \int_0^1 y[2(1-y)]\, dy = 2\int_0^1 (y - y^2)\, dy. \] \[ \bar{y} = 2\Big[\tfrac{y^2}{2} - \tfrac{y^3}{3}\Big]_0^1 = 2(\tfrac{1}{2} - \tfrac{1}{3}) = \tfrac{1}{3}. \]