To evaluate the line integral $\int_C (9y^2x \, dx + 4x^2y \, dy)$ using Green's Theorem, let's break it into steps.

Step 1: Green's Theorem

Green's Theorem states: $\int_C (P \, dx + Q \, dy) = \int\int_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \, dA$ where:

• $C$ is the positively oriented boundary of region $R$,
• $P = 9y^2x$,
• $Q = 4x^2y$.

Step 2: Compute Partial Derivatives

We compute:

• $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x^2y) = 8xy$,
• $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(9y^2x) = 18yx$.

Thus: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 8xy - 18xy = -10xy.$

Step 3: Set Up the Double Integral

The region $R$ is the square with vertices $(0, 0), (1, 0), (1, 1), (0, 1)$. For this square:

• $x \in [0, 1]$,
• $y \in [0, 1]$.

The double integral becomes: $\int\int_R (-10xy) \, dA = \int_0^1 \int_0^1 (-10xy) \, dx \, dy.$

Step 4: Evaluate the Double Integral

First, integrate with respect to $x$: $\int_0^1 (-10xy) \, dx = \int_0^1 -10y \cdot x \, dx = -10y \left[\frac{x^2}{2}\right]_0^1 = -10y \cdot \frac{1}{2} = -5y.$

Now, integrate with respect to $y$: $\int_0^1 (-5y) \, dy = \int_0^1 -5 \cdot y \, dy = -5 \left[\frac{y^2}{2}\right]_0^1 = -5 \cdot \frac{1}{2} = -\frac{5}{2}.$

Final Answer:

The value of the line integral is: $\int_C (9y^2x \, dx + 4x^2y \, dy) = -\frac{5}{2}.$

Would you like additional details or an explanation for any part of this? Here are related questions to consider:

1. What does Green's Theorem imply geometrically?
2. How does the orientation of the curve affect the result?
3. Could this problem be evaluated without Green's Theorem? How?
4. How would the result change if the region $R$ were a triangle instead of a square?
5. What role do partial derivatives play in Green's Theorem?

Tip: Always confirm the orientation (positive or negative) of the curve when applying Green’s Theorem!