Answer
$\dfrac{52}{9}$
Work Step by Step
The domain $D$ for given region can be expressed as: $0 \leq y \leq 4$ and $\sqrt y \leq x \leq 2$
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A= \int_{0}^{4} \int_{\sqrt y}^{2} \sqrt {x^3+1} \ dx \ dy \\= \int_{0}^{2} \int_0^{x^2} \sqrt {x^3+1} dy dx\\= \int_{0}^{2} (y\sqrt {x^3+1})_0^{x^2} \ dx \\= \int_{0}^{2} x^2\sqrt {x^3+1} \ dx $
Suppose that $x^2+1=a \implies x^2 dx=\dfrac{da}{3}$
Now, $ \int_{0}^{2} x^2\sqrt {x^3+1} \ dx = \int_{0}^{2} \dfrac{1}{3} \sqrt t dt\\=[\dfrac{2}{9} a^{3/2}]_0^2 \\=[\dfrac{2}{9} (x^3+1)^{3/2}]_0^2=\dfrac{2}{9} \times (9)^{3/2} \\=\dfrac{52}{9}$
