Answer
$\dfrac{9}{4}$
Work Step by Step
We need to find the points of intersection of the two curves as $y-y^2=y $ and $y^2-y-2=0$
$\implies y=-1, 2$
Now, when $y =2 $, then we have $x=4$
Next, when $y = -1 $, then we have: $x=1$
Thus, the points of intersection can be written as: $(4,2)$ and $(1,-1)$.
We need to express the area of the surface as:
$$\iint_{D} y dA=\int_{-1}^{2} \int_{y^2}^{y+2} y \ dx \ dy \\= \int_{-1}^{2} [yx]_{y^2}^{y+2} dy \\ = \int_{-1}^2 (y^2+2y-y^3) dy \\ = [\dfrac{y^3}{3}+y^2-\dfrac{y^{4}}{4}]_{-1}^2\\=[\dfrac{2^3}{3}+2^2-\dfrac{2^{4}}{4}]-[\dfrac{(-1)^3}{3}+(-1)^2-\dfrac{(-1)^{4}}{4}] \\= \dfrac{8}{3}-\dfrac{5}{12}\\ =\dfrac{9}{4}$$
