Green's Theorem states that for a positively oriented, simple closed curve $C$ enclosing a region $D$, the line integral around $C$ can be converted into a double integral over $D$:

$\oint_C \left( M \, dx + N \, dy \right) = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA$

Given the line integral:

$\oint_C (4y + x) \, dx + (y + x) \, dy$

we identify $M = 4y + x$ and $N = y + x$. We need to find $\frac{\partial N}{\partial x}$ and $\frac{\partial M}{\partial y}$.

1. Calculate $\frac{\partial N}{\partial x}$: $\frac{\partial N}{\partial x} = \frac{\partial (y + x)}{\partial x} = 1$

2. Calculate $\frac{\partial M}{\partial y}$: $\frac{\partial M}{\partial y} = \frac{\partial (4y + x)}{\partial y} = 4$

3. Now, compute $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}$: $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - 4 = -3$

4. Using Green's Theorem, the integral becomes: $\oint_C (4y + x) \, dx + (y + x) \, dy = \iint_D (-3) \, dA$

5. The region $D$ is the interior of the circle $(x - 6)^2 + (y - 6)^2 = 4$. This is a circle with center $(6, 6)$ and radius $2$.

The area $A$ of the circle is: $A = \pi r^2 = \pi \cdot 2^2 = 4\pi$

6. Therefore, the double integral over the region $D$ is: $\iint_D (-3) \, dA = -3 \times \text{Area of } D = -3 \times 4\pi = -12\pi$

Thus, the value of the given line integral is:

$\oint_C (4y + x) \, dx + (y + x) \, dy = -12\pi$

Would you like a deeper explanation on any step, or any further details?

Here are 5 related questions:

1. How is Green's Theorem used to convert a line integral into a double integral?
2. What are the conditions required for Green's Theorem to be applicable?
3. Can Green's Theorem be used for curves that are not circles? How?
4. What is the geometric interpretation of Green's Theorem?
5. How would the integral change if the curve $C$ was traversed in the opposite direction?

Tip: When using Green's Theorem, always ensure the curve is positively oriented (counter-clockwise), and the region is simply connected, with no holes or isolated points.