can you run me through how to answer this question:   Use Green’s theorem to evaluate  $$ \phi2x^3d\text{ }-\text{ }2y^3dx $$ where C is a positively oriented circle of radius $$ \pi $$ centered at the origin

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 $.

---

### **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:
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 6x^2 - (-6y^2) = 6x^2 + 6y^2.
\]

---

### **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:
\[
\iint_R \left( 6x^2 + 6y^2 \right) \, dA = \int_0^{2\pi} \int_0^\pi 6r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^\pi 6r^3 \, dr \, d\theta.
\]

---

### **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 $:**
\[
\int_0^{2\pi} \frac{3\pi^4}{2} \, d\theta = \frac{3\pi^4}{2} \cdot \int_0^{2\pi} 1 \, d\theta = \frac{3\pi^4}{2} \cdot 2\pi = 3\pi^5.
\]

---

### **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!

Use Green’s theorem to evaluate $$ \phi2x^3d\text{ }-\text{ }2y^3dx $$ where C is a positively oriented circle of radius $$ \pi $$ centered at the origin.

Step-by-step solution to evaluate the line integral using Green's theorem for a positively oriented circle of radius π centered at the origin. This problem involves vector calculus concepts, double integrals, and polar coordinates. Perfect for undergraduate-level calculus students.

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