Let's solve this step-by-step using Green's theorem.

Problem Restated:

We are asked to evaluate the line integral
$I = \oint_C \left( y^3 \, dx - 2x^3 \, dy \right),$
where $C$ is the counterclockwise-oriented circle $x^2 + y^2 = a^2$.

Step 1: Applying Green's Theorem

Green's theorem states: $\oint_C \left( M \, dx + N \, dy \right) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA,$ where $C$ is a positively-oriented, simple closed curve, and $R$ is the region enclosed by $C$. In our case:

• $M(x, y) = y^3$
• $N(x, y) = -2x^3$

Step 2: Partial Derivatives

We now compute the required partial derivatives: $\frac{\partial N}{\partial x} = \frac{\partial (-2x^3)}{\partial x} = -6x^2,$ $\frac{\partial M}{\partial y} = \frac{\partial (y^3)}{\partial y} = 3y^2.$

Step 3: Evaluating the Double Integral

Green’s theorem converts the line integral into the following double integral over the disk $x^2 + y^2 \leq a^2$: $I = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = \iint_R \left( -6x^2 - 3y^2 \right) \, dA.$

In polar coordinates:

• $x = r \cos \theta$,
• $y = r \sin \theta$,
• $dA = r \, dr \, d\theta$,
• The region $R$ is described by $0 \leq r \leq a$ and $0 \leq \theta < 2\pi$.

Thus, the integrand becomes: $-6x^2 - 3y^2 = -6r^2 \cos^2 \theta - 3r^2 \sin^2 \theta.$ Factor out $r^2$: $-6r^2 \cos^2 \theta - 3r^2 \sin^2 \theta = -r^2 \left( 6\cos^2 \theta + 3\sin^2 \theta \right).$

Using the given trigonometric identities:
$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}, \quad \sin^2 \theta = \frac{1 - \cos(2\theta)}{2},$ we get: $6 \cos^2 \theta + 3 \sin^2 \theta = 6 \cdot \frac{1 + \cos(2\theta)}{2} + 3 \cdot \frac{1 - \cos(2\theta)}{2}.$

Simplify: