1. **State the problem:** Find the length of the curve defined by $$x = \frac{y^{3/2}}{3} - y^{1/2}$$ from $$y=1$$ to $$y=9$$.
2. **Formula for arc length:** The length $$L$$ of a curve given by $$x = f(y)$$ from $$y=a$$ to $$y=b$$ is
$$
L = \int_a^b \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
$$
3. **Find the derivative:**
$$
x = \frac{y^{3/2}}{3} - y^{1/2} = \frac{y^{3/2}}{3} - y^{1/2}
$$
Calculate $$\frac{dx}{dy}$$:
$$
\frac{dx}{dy} = \frac{3}{2} \cdot \frac{y^{1/2}}{3} - \frac{1}{2} y^{-1/2} = \frac{1}{2} y^{1/2} - \frac{1}{2} y^{-1/2} = \frac{1}{2} \left(y^{1/2} - y^{-1/2}\right)
$$
4. **Square the derivative:**
$$
\left(\frac{dx}{dy}\right)^2 = \left(\frac{1}{2} (y^{1/2} - y^{-1/2})\right)^2 = \frac{1}{4} (y - 2 + y^{-1}) = \frac{y + y^{-1} - 2}{4}
$$
5. **Set up the integral:**
$$
L = \int_1^9 \sqrt{1 + \frac{y + y^{-1} - 2}{4}} \, dy = \int_1^9 \sqrt{\frac{4 + y + y^{-1} - 2}{4}} \, dy = \int_1^9 \sqrt{\frac{y + y^{-1} + 2}{4}} \, dy
$$
6. **Simplify the expression under the square root:**
$$
\frac{y + y^{-1} + 2}{4} = \frac{(\sqrt{y} + \frac{1}{\sqrt{y}})^2}{4} = \left(\frac{\sqrt{y} + \frac{1}{\sqrt{y}}}{2}\right)^2
$$
7. **Therefore, the integrand simplifies to:**
$$
\sqrt{\frac{y + y^{-1} + 2}{4}} = \frac{\sqrt{y} + \frac{1}{\sqrt{y}}}{2}
$$
8. **Rewrite the integral:**
$$
L = \int_1^9 \frac{\sqrt{y} + \frac{1}{\sqrt{y}}}{2} \, dy = \frac{1}{2} \int_1^9 \left(y^{1/2} + y^{-1/2}\right) dy
$$
9. **Integrate term-by-term:**
$$
\int y^{1/2} dy = \frac{2}{3} y^{3/2}, \quad \int y^{-1/2} dy = 2 y^{1/2}
$$
10. **Evaluate the integral:**
$$
L = \frac{1}{2} \left[ \frac{2}{3} y^{3/2} + 2 y^{1/2} \right]_1^9 = \frac{1}{2} \left( \frac{2}{3} (9)^{3/2} + 2 (9)^{1/2} - \frac{2}{3} (1)^{3/2} - 2 (1)^{1/2} \right)
$$
11. **Calculate powers:**
$$
(9)^{1/2} = 3, \quad (9)^{3/2} = (9^{1/2})^3 = 3^3 = 27
$$
12. **Substitute values:**
$$
L = \frac{1}{2} \left( \frac{2}{3} \times 27 + 2 \times 3 - \frac{2}{3} \times 1 - 2 \times 1 \right) = \frac{1}{2} \left( 18 + 6 - \frac{2}{3} - 2 \right)
$$
13. **Simplify inside the parentheses:**
$$
18 + 6 - 2 = 22, \quad 22 - \frac{2}{3} = \frac{66}{3} - \frac{2}{3} = \frac{64}{3}
$$
14. **Final length:**
$$
L = \frac{1}{2} \times \frac{64}{3} = \frac{32}{3} \approx 10.67
$$
**Answer:** The length of the curve from $$y=1$$ to $$y=9$$ is $$\boxed{\frac{32}{3}}$$ or approximately 10.67 units.
Curve Length A91E6D
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