To evaluate the given line integral using Green's Theorem, follow these steps:

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Step 1: Recall Green's Theorem

Green's theorem states: $\oint_C P \, dx + Q \, dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$ where:

• $C$ is a positively oriented, simple closed curve enclosing a region $R$,
• $P$ and $Q$ are functions of $x$ and $y$.

In this problem:

• $P(x, y) = -2y^3$,
• $Q(x, y) = 2x^3$.

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Step 2: Identify the Partial Derivatives

Compute the necessary derivatives:

1. $\frac{\partial Q}{\partial x}$: Derivative of $Q = 2x^3$ with respect to $x$, $\frac{\partial Q}{\partial x} = 6x^2.$
2. $\frac{\partial P}{\partial y}$: Derivative of $P = -2y^3$ with respect to $y$, $\frac{\partial P}{\partial y} = -6y^2.$

Substitute into Green's theorem: $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x^2 - (-6y^2) = 6x^2 + 6y^2.$

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Step 3: Set Up the Double Integral

The region $R$ is a circle of radius $\pi$ centered at the origin. In polar coordinates:

• $x = r\cos\theta$,
• $y = r\sin\theta$,
• $dA = r \, dr \, d\theta$.

The bounds for $r$ and $\theta$ are: $0 \leq r \leq \pi, \quad 0 \leq \theta \leq 2\pi.$

The integrand $6x^2 + 6y^2$ becomes: 6(x^2 + y^2) = 6r^2 \quad \text{(since x^2 + y^2 = r^2)}.

Thus, the integral becomes: $\iint_R \left( 6x^2 + 6y^2 \right) \, dA = \int_0^{2\pi} \int_0^\pi 6r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^\pi 6r^3 \, dr \, d\theta.$

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Step 4: Evaluate the Integral

1. Inner integral over $r$: $\int_0^\pi 6r^3 \, dr = 6 \int_0^\pi r^3 \, dr = 6 \left[ \frac{r^4}{4} \right]_0^\pi = 6 \cdot \frac{\pi^4}{4} = \frac{3\pi^4}{2}.$

2. Outer integral over $\theta$: $\int_0^{2\pi} \frac{3\pi^4}{2} \, d\theta = \frac{3\pi^4}{2} \cdot \int_0^{2\pi} 1 \, d\theta = \frac{3\pi^4}{2} \cdot 2\pi = 3\pi^5.$

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Final Answer:

$\oint_C (2x^3 \, dy - 2y^3 \, dx) = 3\pi^5$

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Do you have any questions or would you like further clarification?

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5 Related Questions:

1. What are the conditions for applying Green’s theorem to a line integral?
2. How can Green's theorem be used to compute the area enclosed by a curve?
3. How do you convert a double integral to polar coordinates?
4. How would the solution change if the region was an ellipse instead of a circle?
5. Can Green’s theorem be used for non-conservative vector fields?

Tip:

When applying Green's theorem, always ensure the curve is positively oriented (counterclockwise) to avoid sign errors!