Answer
$$\int_{-4}^{4} \int_{0}^{x^{2}} \sqrt{64-x^{3}} d y d x=\frac{2048}{9} \sqrt{2}$$
Work Step by Step
Given $$\int_{-4}^{4} \int_{0}^{x^{2}} \sqrt{64-x^{3}} d y d x$$
So, we get
\begin{align}
\int_{-4}^{4} \int_{0}^{x^{2}} \sqrt{64-x^{3}} d y \ d x&=\int_{-4}^{4}\left[y \sqrt{64-x^{3}}\right]_{0}^{2} d x\\
&=\int_{-4}^{4} x^{2} \sqrt{64-x^{3}} d x\\
&=\left[-\frac{2}{9}\left(64-x^{3}\right)^{3 / 2}\right]_{-4}^{4}\\
&=0+\frac{2}{9}(128)^{3 / 2}\\
&=\frac{2048}{9} \sqrt{2}
\end{align}
