Green's Theorem relates a line integral around a closed curve $C$ to a double integral over the region $R$ enclosed by $C$. The theorem states:

$\oint_C \left( P \, dx + Q \, dy \right) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy$

In this case, the double integral we are given is $\iint_R y^2 \, dx \, dy$. This can be evaluated using Green's Theorem by appropriately choosing the vector field $\mathbf{F} = P \, \hat{i} + Q \, \hat{j}$.

We need to find a vector field $\mathbf{F}$ such that:

$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y^2$

Step 1: Choosing $P$ and $Q$

We aim for $\frac{\partial Q}{\partial x} = y^2$ and $\frac{\partial P}{\partial y} = 0$. Therefore, we can choose:

• $P(x, y) = 0$
• $Q(x, y) = \frac{1}{3} y^3$

This satisfies $\frac{\partial Q}{\partial x} = y^2$ and $\frac{\partial P}{\partial y} = 0$.

Step 2: Applying Green's Theorem

Using Green's Theorem, we convert the double integral into a line integral over the boundary of the region $R$ (which is the boundary of the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$):

$\iint_R y^2 \, dx \, dy = \oint_C \left( P \, dx + Q \, dy \right) = \oint_C \left( 0 \, dx + \frac{1}{3} y^3 \, dy \right)$

Thus, the double integral $\iint_R y^2 \, dx \, dy$ is transformed into the line integral:

$\oint_C \frac{1}{3} y^3 \, dy$

Step 3: Parameterizing the Ellipse

The boundary $C$ of the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$ can be parameterized using the parametric equations for an ellipse:

$x = 4 \cos t, \quad y = 2 \sin t \quad \text{for} \quad t \in [0, 2\pi]$

Now, we can compute the line integral in terms of $t$:

$\oint_C \frac{1}{3} y^3 \, dy = \int_0^{2\pi} \frac{1}{3} (2 \sin t)^3 \cdot 2 \cos t \, dt$

Simplifying:

$= \frac{8}{3} \int_0^{2\pi} \sin^3 t \cos t \, dt$

Step 4: Simplifying the Integral

We use the identity $\sin^3 t = (3 \sin t - \sin 3t)/4$ to simplify the integral:

$\frac{8}{3} \int_0^{2\pi} \left( \frac{3 \sin t - \sin 3t}{4} \right) \cos t \, dt$

Each term of this integral can be evaluated using standard trigonometric integrals, leading to the final result.

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Would you like further details or to proceed with solving this integral step-by-step?

Here are 5 follow-up questions that extend this topic:

1. How does Green's Theorem relate to Stokes' Theorem in vector calculus?
2. What is the geometric interpretation of Green's Theorem in the plane?
3. How would you modify the parameterization for a general ellipse with axes $a$ and $b$?
4. How does the boundary orientation (clockwise or counterclockwise) affect the sign of the line integral?
5. How can Green’s Theorem be applied to compute the area enclosed by a curve?

Tip: When applying Green's Theorem, it’s crucial to correctly orient the boundary curve $C$; it must follow the positive (counterclockwise) orientation.