Answer
$$\frac{94\pi}{225}$$
Work Step by Step
Given $$ y=\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}, \quad[0,1]$$
\begin{align*}
y'&= \frac{1}{2} x^{-1 / 4}-\frac{1}{2} x^{1 / 4}\\
1+y'^2&= 1+\left(\frac{1}{2} x^{-1 / 4}-\frac{1}{2} x^{1 / 4}\right)^2\\
&=\left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right)^2
\end{align*}
Then the surface area is given by
\begin{align*}
S&=2\pi \int_{a}^{b} y\sqrt{1+y'^2}dx\\
&=2\pi \int_{0}^{1}\left(\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}\right)\sqrt{\left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right)^2} dx\\
&=2\pi \int_{0}^{1}\left(\frac{2}{3} x^{3 / 4}-\frac{2}{5} x^{5 / 4}\right) \left(\frac{1}{2} x^{-1 / 4}+\frac{1}{2} x^{1 /4}\right) dx\\
&=2\pi \int_0^1 [\frac{1}{3} x^{1/2} +\frac{1}{3}x -\frac{1}{5} x -\frac{1}{5} x^{3/2}]dx\\
&=2\pi[ \frac{2}{9} x^{3/2} +\frac{1}{15}x^2 -\frac{2}{25} x^{5/2 }]\bigg|_{0}^{1}\\
&=\frac{94\pi}{225}
\end{align*}
