Answer
$\dfrac{2}{3}$
Work Step by Step
We can express the region $D$ in the triangular region as follows:
$$D=\left\{ (x, y) | 0 \leq y \leq 1, \ y \leq x \leq 4-3y \right\}
$$
Therefore, $$\iint_{D} (y) \ dA=\int_{0}^{1} \int_{x }^{4-3y} y \ dx \ dy \\ =\int_{0}^{1} [xy]_{ x}^{4-3y} \ dy \\= \int_{0}^{1} [y(4-3y) -y^2] dy \\ = \int_{0}^1 [4y -4y^2] dy=[2y^2-\dfrac{4y^3}{3}]_0^1 \\= 2-\dfrac{4}{3} \\= \dfrac{2}{3}$$
