Answer
$\dfrac{4 \sqrt 2 -2}{9}$
Work Step by Step
We can define the domain $D$ in the Type-1 using vertical cross-sections as follows: $
D=\left\{ (x, y) | \sqrt x \leq y \leq 1, \ 0 \leq x \leq 1 \right\}
$
Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{\sqrt x}^{1} \sqrt {y^3 +1} \ dy \ dx $
and we can define the domain $D$ in the Type-II using horizontal cross-sections as follows:
$D=\left\{ (x, y) | 0 \leq x \leq y^2, \ 0 \leq y \leq 1 \right\}
$
Therefore, $\iint_{D} f(x,y) dA=\int_{0}^{1} \int_{0}^{y^2} \sqrt {y^3 +1} \ dx \ dy \\=\int_0^1 [x \sqrt {y^3 +1} ]_0^{y^2} \ dy \\=\int_0^1 y^2 \sqrt {y^3+1} dy$
Let us consider $a=y^3+1 \implies da=3y^2 dy$
Now, $\int_0^1 y^2 \sqrt {y^3+1} dy=\int_1^2 \sqrt a \dfrac{da}{3} \\ =\int_1^2 \dfrac{a^{1/2}}{3} da \\= [\dfrac{2 a \sqrt a}{9}]_1^2\\= [\dfrac{2}{9} \times 2 \times \sqrt 2 - \dfrac{2}{9} \times 1 \times \sqrt 1 ]\\=\dfrac{4 \sqrt 2 -2}{9}$
