Answer
$$\frac{67}{36} \pi$$
Work Step by Step
Given $$y=\frac{2}{3} x^{3 / 2}-\frac{1}{2} x^{1 / 2}, \quad[1,2]$$
\begin{align*}
y'&= \sqrt{x}-\frac{1}{4 \sqrt{x}}\\
1+y'^2&= 1+\left(\sqrt{x}-\frac{1}{4 \sqrt{x}}\right)^{2}\\
&=1+\left(x-\frac{1}{2}+\frac{1}{16 x}\right)\\
&=x+\frac{1}{2}+\frac{1}{16 x}\\
&=\left(\sqrt{x}+\frac{1}{4 \sqrt{x}}\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_{1}^{2}\left(\frac{2}{3} x^{3 / 2}-\frac{\sqrt{x}}{2}\right)\left(\sqrt{x}+\frac{1}{4 \sqrt{x}}\right) d x\\
&=2 \pi \int_{1}^{2}\left(\frac{2}{3} x^{2}+\frac{1}{6} x-\frac{1}{2} x-\frac{1}{8}\right) d x\\
&=\left.2 \pi\left(\frac{2 x^{3}}{9}-\frac{x^{2}}{6}-\frac{1}{8} x\right)\right|_{1} ^{2}\\
&=\frac{67}{36} \pi
\end{align*}
