Answer
$\dfrac{9}{4}$
Work Step by Step
We will compute the points of intersection of the two curves as follows: $y-y^2=y \implies y^2-y-2=0$
Thus, $y=-1, 2$
So, when $y =2 $, then $x=4$ and $y = -1 $, then $x=1$
Therefore, the points of intersection are: $(4,2)$ and $(1,-1)$.
We can define the area of the surface as:
$\iint_{D} y dA=\int_{-1}^{2} \int_{y^2}^{y+2} y \ dx \ dy \\= \int_{-1}^{2} [yx]_{y^2}{y+2} dy \\ = \int_{-1}^2 (y^2+2y-y^3) dy \\ = [\dfrac{y^3}{3}+y^2-\dfrac{y^{4}}{4}]_{-1}^2 \\= \dfrac{8}{3}-\dfrac{5}{12}\\ =\dfrac{9}{4}$
