Answer
\[
\frac{-\cos 8+1}{3}
\]
Work Step by Step
Using the rules of integrals, we obtain:
\[
\begin{array}{c}
\int_{0}^{2} \int_{0}^{y^{2}} \sin y^{3} d x d y=\iint \sin y^{3} d A \\
=\int_{0}^{2}\left[x \sin y^{3}\right]_{0}^{y^{2}} d y \\
=\int_{0}^{2} y^{2} \sin y^{3} d y \\
=\frac{1}{3} \int_{0}^{2} 3 y^{2} \sin y^{3} d y
\end{array}
\]
Replace $3 y^{2} d y=d u$ and $u=y^{3}$
$=\frac{1}{3} \int_{0}^{8} \sin u d u$
\[
\begin{array}{c}
=\frac{1}{3}[-\cos u]_{0}^{8} \\
=\frac{-\cos 8+1}{3}
\end{array}
\]
