Answer
$\dfrac{2}{3}$
Work Step by Step
Divergence Theorem: It states that the flux across the boundary of $R$ is equal to the net contraction or expansion of the vector field $F$ within the region $R$.This implies that
$\int\int F \cdot n d C=\int\int\int \nabla \cdot F dR$
Here, $R$ defines the connected and region in space which is oriented by boundary $C$ and n denotes the outward flux unit normal vector on boundary $C$.
In order to compute the net outward flux of the given vector field, we will firstly compute $\nabla \cdot F$.
Now, we have:
$\nabla \cdot F=\dfrac{\partial}{\partial x}(x) +\dfrac{\partial}{\partial y}(2y)+\dfrac{\partial}{\partial y}(z)\\=1+2+1\\=4$
Thus, $\int\int\int \nabla \cdot F dR\\= 4 \iiint \ d V\\=2 \int_0^1 \int_0^{1-x} \int_0^{1-x-y} \ dz \ dy \ dx\\=4 \int_0^1 \int_0^{1-x}(1-x-y)\ dy \ dx \\=4 \int_0^1 [y-yx+\dfrac{y^2}{2}]_0^{1-x} \ dz\\=2 \int_0^1 [1-2x+x^2] \ dx \\=2 [x-\dfrac{2x^2}{2}+\dfrac{x^3}{3}]_0^1\\=2 [1-\dfrac{2(1)^2}{2}+\dfrac{(1)^3}{3}]\\=\dfrac{2}{3}$
