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We are tasked with evaluating the line integral \( I = \oint_C \left( y^3 \, dx - 2x^3 \, dy \right) \), where \( C \) is the counterclockwise circle \( x^2 + y^2 = a^2 \), by using Green's Theorem.

### Step 1: Green's Theorem
Green's theorem relates a line integral around a simple, closed curve \( C \) to a double integral over the region \( R \) enclosed by \( C \). The theorem states:
\[
\oint_C \left( P \, dx + Q \, dy \right) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA
\]
For our problem:
- \( P(x, y) = y^3 \)
- \( Q(x, y) = -2x^3 \)

Thus, we need to compute \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).

### Step 2: Compute the Partial Derivatives
- \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (-2x^3) = -6x^2 \)
- \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (y^3) = 3y^2 \)

Thus, Green's theorem reduces the line integral to the double integral:
\[
\iint_R \left( -6x^2 - 3y^2 \right) \, dA
\]
This can be rewritten as:
\[
\iint_R -3(2x^2 + y^2) \, dA
\]

### Step 3: Convert to Polar Coordinates
To evaluate this integral, it is convenient to switch to polar coordinates, where:
- \( x = r\cos(\theta) \)
- \( y = r\sin(\theta) \)
- \( dA = r \, dr \, d\theta \)
- The equation of the circle \( x^2 + y^2 = a^2 \) becomes \( r = a \), so \( r \in [0, a] \) and \( \theta \in [0, 2\pi] \).

In polar coordinates, we have:
\[
2x^2 + y^2 = 2r^2\cos^2(\theta) + r^2\sin^2(\theta) = r^2(2\cos^2(\theta) + \sin^2(\theta)) = r^2(1 + \cos^2(\theta))
\]

Thus, the integral becomes:
\[
I = \iint_R -3r^2(1 + \cos^2(\theta)) \, r \, dr \, d\theta = -3 \int_0^{2\pi} \int_0^a r^3(1 + \cos^2(\theta)) \, dr \, d\theta
\]

### Step 4: Perform the Integration
1. Integrating with respect to \( r \):
\[
\int_0^a r^3 \, dr = \frac{a^4}{4}
\]
Thus, the integral becomes:
\[
I = -3 \cdot \frac{a^4}{4} \int_0^{2\pi} (1 + \cos^2(\theta)) \, d\theta
\]

2. Now, integrate with respect to \( \theta \). Using the hint from the problem, \( \cos^2(\theta) \) has the identity:
\[
\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}
\]
Thus:
\[
\int_0^{2\pi} \left( 1 + \frac{1 + \cos(2\theta)}{2} \right) \, d\theta = \int_0^{2\pi} \frac{3}{2} \, d\theta = \frac{3}{2} \cdot 2\pi = 3\pi
\]

### Step 5: Final Answer
Substituting this result back into the integral:
\[
I = -3 \cdot \frac{a^4}{4} \cdot 3\pi = -\frac{9\pi a^4}{4}
\]

Thus, the value of the line integral is:
\[
I = -\frac{9\pi a^4}{4}
\]

Let me know if you would like additional details or clarifications!

### Relative questions:
1. What is the main application of Green's Theorem in vector calculus?
2. How do you convert Cartesian coordinates to polar coordinates?
3. Why is it useful to apply Green's Theorem instead of directly calculating line integrals?
4. Can you explain how the polar form simplifies integration over circular regions?
5. What other types of problems can Green's Theorem help solve?

### Tip:
When applying Green's Theorem, always check if converting the problem to polar coordinates simplifies the integrals, especially for regions like circles or disks.

By using Green's theorem, evaluate the curve integral I = ∮C (y^3 dx - 2x^3 dy), where C is the counterclockwise circle x^2 + y^2 = a^2.

Learn how to evaluate a line integral using Green's theorem for a curve defined by the counterclockwise circle x^2 + y^2 = a^2. This solution involves converting the integral into polar coordinates and applying Green's theorem. Suitable for college-level students.

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