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To find the length of the curve defined by \( x = \frac{1}{3} \sqrt{y(y - 3)} \) over the interval \( 1 \leq y \leq 9 \), we will use the formula for arc length when the function is given as \( x = g(y) \):

\[
L = \int_{a}^{b} \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy
\]

### Step 1: Differentiate \( x \) with respect to \( y \).
The given function is:
\[
x = \frac{1}{3} \sqrt{y(y - 3)}
\]

We need to find \(\frac{dx}{dy}\):

\[
\frac{dx}{dy} = \frac{1}{3} \cdot \frac{1}{2\sqrt{y(y - 3)}} \cdot \left(2y - 3\right) = \frac{2y - 3}{6 \sqrt{y(y - 3)}}
\]

### Step 2: Compute the integrand.
Substitute \(\frac{dx}{dy}\) into the arc length formula:

\[
\left(\frac{dx}{dy}\right)^2 = \left(\frac{2y - 3}{6 \sqrt{y(y - 3)}}\right)^2 = \frac{(2y - 3)^2}{36 y(y - 3)}
\]

Thus, the integrand becomes:

\[
\sqrt{1 + \frac{(2y - 3)^2}{36 y(y - 3)}}
\]

### Step 3: Simplify the integrand.
Combining the terms under a common denominator, the integrand simplifies to:

\[
\sqrt{\frac{36 y(y - 3) + (2y - 3)^2}{36 y(y - 3)}}
\]

### Step 4: Evaluate the integral.
The arc length is given by:

\[
L = \int_{1}^{9} \sqrt{\frac{36 y(y - 3) + (2y - 3)^2}{36 y(y - 3)}} \, dy
\]

We'll perform this integration numerically.

Find the length of the curve x = (1/3) * sqrt(y(y-3)) for 1 ≤ y ≤ 9.

Calculate the arc length of the curve defined by x = (1/3) * sqrt(y(y-3)) for the interval 1 ≤ y ≤ 9. The solution uses differentiation, the arc length formula, and integration to find the length. This problem is suitable for high school and college-level calculus students.

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