Answer
$$\frac{-1+e^{8}}{3}$$
Work Step by Step
\[
\iint_{D} e^{x^{3}} d A=\int_{0}^{4} \int_{\sqrt{y}}^{2} e^{x^{3}} d x d y
\]
$D$ has been interpreted as a type II region (using horizontal cross-sections).
We can also Interpret $D$ as a type I region (using vertical cross-sections).
So:
\[
\begin{array}{l}
\int_{0}^{2} \int_{0}^{x^{2}} e^{x^{3}} d y d x=\iint_{D} e^{x^{3}} d A \\
\quad=\int_{0}^{2}\left[y e^{x^{3}}\right]_{y=0}^{y=x^{2}} d x \\
\quad=\int_{0}^{2} x^{2} e^{x^{3}} d x
\end{array}
\]
Replace $d u=3 x^{2} d x$ and $u=x^{3}$
\[
=\frac{1}{3} \int_{0}^{8} e^{u} d u
\]
\[
\begin{array}{l}
=\frac{1}{3}\left[e^{u}\right]_{0}^{8} \\
=\frac{-1+e^{8}}{3}
\end{array}
\]
