Answer
$$\frac{384 \pi}{5}$$
Work Step by Step
Since
$$
1+\left(y^{\prime}\right)^{2}=1+\left(-\frac{\sqrt{4-x^{2 / 3}}}{\sqrt[3]{x}}\right)^{2}=1+\frac{4-x^{2 / 3}}{x^{2 / 3}}=\frac{4}{x^{2 / 3}}
$$
Then
\begin{aligned}
\text { Surface Area } &=2 \pi \int_{0}^{8}\left(4-x^{2 / 3}\right)^{3 / 2} \sqrt{\frac{4}{x^{2 / 3}}} d x \\
&=2 \pi \int_{0}^{8}\left(4-x^{2 / 3}\right)^{3 / 2} \frac{2}{x^{1 / 3}} d x \\
&=4 \pi \int_{0}^{8} \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x
\end{aligned}
Let $$
u=4-x^{2 / 3} \longrightarrow d x=-\frac{3 \sqrt[3]{x}}{2}
$$
Then
\begin{aligned}
\int \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x &=-\frac{3}{2} \int \frac{u^{3 / 2} \sqrt[3]{x}}{\sqrt[3]{x}} d u \\ &=-\frac{3}{2} \int u^{3 / 2} d u \\ &=-\frac{3}{2}\left[\frac{2}{5} u^{5 / 2}\right] \\
& =-\frac{3}{5}\left(4-x^{2 / 3}\right)^{5 / 2}
\end{aligned}
Hence
\begin{aligned}
\text { Surface Area }&=4 \pi \int_{0}^{8} \frac{\left(4-x^{2 / 3}\right)^{3 / 2}}{\sqrt[3]{x}} d x &=4 \pi\left[-\frac{3}{5}\left(4-x^{2 / 3}\right)^{5 / 2}\right]_{0}^{8} \\
&=4 \pi\left[\frac{96}{5}\right] \\
&=\frac{384 \pi}{5}
\end{aligned}
