To solve this problem, we will use Green's Theorem, which relates a line integral around a simple closed curve $C$ to a double integral over the region $R$ enclosed by $C$. Green's Theorem states:

$\oint_{C} \left( P \, dx + Q \, dy \right) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$

In this case, we are given: $P = e^x - 3y^2 \quad \text{and} \quad Q = x^2$

Step 1: Compute the partial derivatives

To apply Green's Theorem, we need to calculate $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$.

1. Calculate $\frac{\partial Q}{\partial x}$: $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x$

2. Calculate $\frac{\partial P}{\partial y}$: $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(e^x - 3y^2) = -6y$

Step 2: Set up the double integral

According to Green's Theorem: $\oint_{C} \left( P \, dx + Q \, dy \right) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA$ Substitute the values we found: $\iint_{R} \left( 2x + 6y \right) \, dA$

Step 3: Define the region $R$

The curve $C$ is a semicircle of radius 2, centered at the origin, covering the upper half of the circle (since it’s a semicircle). In polar coordinates, the region $R$ is described by: $0 \leq r \leq 2, \quad 0 \leq \theta \leq \pi$

Step 4: Convert to polar coordinates

In polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$. The area element $dA$ in polar coordinates is $r \, dr \, d\theta$.

Thus, $2x + 6y = 2(r \cos \theta) + 6(r \sin \theta) = r(2 \cos \theta + 6 \sin \theta)$

The integral becomes: $\iint_{R} \left( r(2 \cos \theta + 6 \sin \theta) \right) \, r \, dr \, d\theta = \int_0^{\pi} \int_0^2 r^2 (2 \cos \theta + 6 \sin \theta) \, dr \, d\theta$

Step 5: Evaluate the integral

1. Integrate with respect to $r$: $\int_0^2 r^2 (2 \cos \theta + 6 \sin \theta) \, dr = \left[ \frac{r^3}{3} (2 \cos \theta + 6 \sin \theta) \right]_0^2 = \frac{8}{3} (2 \cos \theta + 6 \sin \theta)$

2. Integrate with respect to $\theta$: $\int_0^{\pi} \frac{8}{3} (2 \cos \theta + 6 \sin \theta) \, d\theta = \frac{8}{3} \left( \int_0^{\pi} 2 \cos \theta \, d\theta + \int_0^{\pi} 6 \sin \theta \, d\theta \right)$

Evaluating each part:

• For $\int_0^{\pi} 2 \cos \theta \, d\theta = 0$ because the integral of $\cos \theta$ over $[0, \pi]$ is zero.
• For $\int_0^{\pi} 6 \sin \theta \, d\theta$: $\int_0^{\pi} 6 \sin \theta \, d\theta = -6 \cos \theta \Big|_0^{\pi} = -6(-1 - 1) = 12$

So, $\frac{8}{3} \cdot 12 = 32$

Final Answer

The value of the line integral is approximately $\boxed{32}$.

Would you like further details on any of the steps?

Here are some follow-up questions for additional exploration:

1. How does Green's Theorem simplify calculations of line integrals over closed curves?
2. What are the conditions required for Green’s Theorem to be applicable?
3. How would the answer change if the region $R$ was a full circle instead of a semicircle?
4. Can Green’s Theorem be extended to three dimensions, and what is that extension called?
5. How does the use of polar coordinates benefit the evaluation of this integral?

Tip: When converting to polar coordinates, always ensure the limits reflect the geometry of the region correctly, as this can significantly simplify integration.