Answer
$$6a$$
Work Step by Step
Since
\begin{aligned}
y^{\prime}(x) &=\frac{d}{d x}\left[\left(a^{2 / 3}-x^{2 / 3}\right)^{3 / 2}\right] \\
&=\frac{3}{2}\left(a^{2 / 3}-x^{2 / 3}\right)^{1 / 2} \cdot\left(0-\frac{2}{3} x^{-1 / 3}\right) \\
&=\frac{-\left(a^{2 / 3}-x^{2 / 3}\right)^{1 / 2}}{x^{1 / 3}}
\end{aligned}
Then
\begin{align*}
s&=4 \int_{0}^{a} \sqrt{1+\left[\frac{-\left(a^{2 / 3}-x^{2 / 3}\right)^{1 / 2}}{x^{1 / 3}}\right]^{2}} d x\\
&==4 \int_{0}^{a} \sqrt{1+\frac{a^{2 / 3}-x^{2 / 3}}{x^{2 / 3}}} d x\\
&=4 \int_{0}^{a} \sqrt{\frac{x^{2 / 3}+a^{2 / 3}-x^{2 / 3}}{x^{2 / 3}}} d x\\
&=4 \int_{0}^{a} \sqrt{\frac{x^{2 / 3}+a^{2 / 3}-x^{2 / 3}}{x^{2 / 3}}} d x\\
&=4 \int_{0}^{a} \sqrt{\frac{a^{2 / 3}}{x^{2 / 3}}} d x\\
&=4 \int_{0}^{a} a^{1 / 3} x^{-1 / 3} d x\\
&=4 a^{1 / 3} \cdot \frac{3 a^{2 / 3}}{2 / 3}\\
&=6a
\end{align*}
