Answer
$\dfrac{11}{3}$
Work Step by Step
Our aim is to calculate the area of the given surface.
The domain $D$ in Type-1 can be expressed as follows:
$D=\left\{ (x, y) | 1 \leq y \leq 2 , \ y-1 \leq x \leq -3y+7 \right\}
$
Now, $A=\iint_{D} y^2 \ dA=\int_{1}^{2} \int_{y-1}^{-3y+7} y^2 \ dx \ dy \\ =\int_{1}^{2} [xy^2]_{y-1}^{-3y+7} \ dx$
and $A= \int_1^2 (-4y^3+8y^2) \ dy \\= [\dfrac{-4y^4}{4} +\dfrac{8y^3}{3}]_1^2 \\= [-y^4+\dfrac{8y^3}{3}]_1^2\\ [-(2^4-(1)^4]-(8/3)[2^3 -1^3] \\=-15+\dfrac{56}{3} \\=\dfrac{11}{3}$
