The given problem involves evaluating the line integral along a curve $C$ bounded by the curves $y = x^3$ and $y = \sqrt{x}$. The integral to be evaluated is:

$\int_C \left( x^2 y \, dx - 9x^2 y \, dy \right)$

with the region bounded by $y = x^3$ and $y = \sqrt{x}$.

Steps to solve:

1. Set the bounds of integration: The two curves given are $y = x^3$ and $y = \sqrt{x}$. To determine the bounds of integration, we find the points where these two curves intersect by solving $x^3 = \sqrt{x}$:

   $x^3 = \sqrt{x} \quad \text{or} \quad x^6 = x \quad \Rightarrow \quad x^6 - x = 0 \quad \Rightarrow \quad x(x^5 - 1) = 0$

   This gives the solutions $x = 0$ and $x = 1$. So, the region is bounded between $x = 0$ and $x = 1$.

2. Parameterize the curves:

   • For $y = x^3$, the parameterization is $x = t$, $y = t^3$.
   • For $y = \sqrt{x}$, the parameterization is $x = t$, $y = \sqrt{t}$.

3. Evaluate the line integral: The line integral is evaluated over the two curves $y = x^3$ and $y = \sqrt{x}$, and we need to handle each curve separately.

   For $y = x^3$ (lower curve, $x$ from 0 to 1):

   • $x = t$, $y = t^3$
   • $dx = dt$, $dy = 3t^2 dt$
   • The integral becomes: $\int_0^1 \left( t^2 (t^3) \, dt - 9t^2 (t^3) \cdot 3t^2 \, dt \right)$ Simplifying: $\int_0^1 \left( t^5 \, dt - 27t^7 \, dt \right) = \left[ \frac{t^6}{6} - \frac{27t^8}{8} \right]_0^1 = \frac{1}{6} - \frac{27}{8}$ $= \frac{1}{6} - \frac{27}{8} = \frac{4 - 81}{24} = \frac{-77}{24}$

   For $y = \sqrt{x}$ (upper curve, $x$ from 1 to 0):

   • $x = t$, $y = \sqrt{t}$
   • $dx = dt$, $dy = \frac{1}{2\sqrt{t}} dt$
   • The integral becomes: $\int_1^0 \left( t^2 (\sqrt{t}) \, dt - 9t^2 (\sqrt{t}) \cdot \frac{1}{2\sqrt{t}} \, dt \right)$ Simplifying: $\int_1^0 \left( t^{5/2} \, dt - \frac{9}{2}t^2 \, dt \right)$ Change the limits and compute: $\int_0^1 \left( t^{5/2} \, dt - \frac{9}{2} t^2 \, dt \right) = \left[ \frac{t^{7/2}}{\frac{7}{2}} - \frac{9}{2} \cdot \frac{t^3}{3} \right]_0^1$ $= \frac{2}{7} - \frac{9}{6} = \frac{2}{7} - \frac{3}{2} = \frac{4 - 21}{14} = \frac{-17}{14}$

4. Add the two results: The total integral is: $\frac{-77}{24} + \frac{-17}{14}$ Simplifying: $\frac{-77}{24} + \frac{-17}{14} = \frac{-539}{168} + \frac{-204}{168} = \frac{-743}{168}$

Thus, the value of the integral is $\frac{-743}{168}$.

Would you like further details on any part of the solution? Here are some related questions:

1. How do we determine the orientation for the line integral?
2. What is the significance of parameterizing the curves for the integral?
3. Could this problem be solved using Green's Theorem?
4. What happens if we reverse the limits of integration for one of the curves?
5. How do the $dx$ and $dy$ terms affect the final integral calculation?

Tip: Always check for symmetry or simplifications in the curves and the integral when evaluating complex line integrals.