Answer
$\dfrac{2}{3}$
Work Step by Step
Our aim is to calculate the area of the given surface.
The domain $D$ can be expressed as follows:
$D=\left\{ (x, y) | 0 \leq y \leq 1 , \ y \leq x \leq 4-3y \right\}
$
Now, $\iint_{D} (y) \ dA=\int_{0}^{1} \int_{x }^{4-3y} y \ dx \ dy \\ =\int_{0}^{1} [xy]_{ x}^{4-3y} \ dy \\= \int_{0}^{1} [y(4-3y) -y^2] dy \\ = \int_{0}^1 [4y -3y^2 -y^2] dy \\=[2y^2]_0^1 -[ \dfrac{4y^3}{3}]_0^1 \\= \dfrac{6-4}{3} \\= \dfrac{2}{3}$
