Answer
The exact length of the curve is:
$$
x=\frac{1}{3} \sqrt{y}(y-3), 1 \leq y \leq9
$$
is equal to $ \frac{32}{3}. $
Work Step by Step
$$
x=\frac{1}{3} \sqrt{y}(y-3), 1 \leq y \leq9.
$$
We have
$$
x=\frac{1}{3} \sqrt{y}(y-3)=\frac{1}{3} y^{3 / 2}-y^{1 / 2}
$$
$\Rightarrow$
$$
d x / d y=\frac{1}{2} y^{1 / 2}-\frac{1}{2} y^{-1 / 2} \Rightarrow
$$
$$
\begin{aligned} 1+(d x / d y)^{2} &=1+\frac{1}{4} y-\frac{1}{2}+\frac{1}{4} y^{-1} \\
& =\frac{1}{4} y+\frac{1}{2}+\frac{1}{4} y^{-1} \\
& =\left(\frac{1}{2} y^{1 / 2}+\frac{1}{2} y^{-1 / 2}\right)^{2}.
\end{aligned}
$$
So the arc length is
$$
\begin{aligned} L &=\int_{1}^{9}\left(\frac{1}{2} y^{1 / 2}+\frac{1}{2} y^{-1 / 2}\right) d y\\
&=\frac{1}{2}\left[\frac{2}{3} y^{3 / 2}+2 y^{1 / 2}\right]_{1}^{9}\\
&=\frac{1}{2}\left[\left(\frac{2}{3} \cdot 27+2 \cdot 3\right)-\left(\frac{2}{3} \cdot 1+2 \cdot 1\right)\right] \\
& = \frac{1}{2}\left(24-\frac{8}{3}\right)=\frac{1}{2}\left(\frac{64}{3}\right)\\
&=\frac{32}{3}.
\end{aligned}
$$
Hence, the exact length of the curve
$$
x=\frac{1}{3} \sqrt{y}(y-3), 1 \leq y \leq9
$$
is equal to $ \frac{32}{3}. $
