Answer
$\dfrac{9}{4}$
Work Step by Step
The points of intersection for the two curves can be defined
as follows: $y-y^2=y $ or, $y^2-y-2=0$
So, $\implies y=-1, 2$
Consider $y =2 $ , then $x=4$
Now, set $y = -1 $ , then $x=1$
Therefore, the points of intersection are: $(4,2)$ and $(1,-1)$.
Now, we can work out with the given integral as follows: $$\iint_{D} y dA=\int_{-1}^{2} \int_{y^2}^{y+2} y \ dx \ dy \\= \int_{-1}^{2} [yx]_{y^2}^{y+2} dy \\ = \int_{-1}^2 (y^2+2y-y^3) dy \\ = [\dfrac{y^3}{3}+y^2-\dfrac{y^{4}}{4}]_{-1}^2\\=\dfrac{(2)^3-(-1)^3}{3}+[(2)^2-(-1)^2]-[\dfrac{2^{4}-(-1)^4}{4}] \\= \dfrac{8}{3}-\dfrac{5}{12}\\ =\dfrac{9}{4}$$
