Answer
$$-24 \pi$$
Work Step by Step
Green's Theorem states that:
$\oint_C A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
We need to set up the line integral and compute the integrand of the double integral as follows:
$\oint_C [-3x^2-3y^2] dA=\iint_{D}(\dfrac{\partial (-3x^2)}{\partial x}-\dfrac{\partial (3y^2) }{\partial y})dA$
Now, convert it to polar coordinates as follows:
$$\oint_C [-3x^2-3y^2] dA=(-3) \times \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times r dr d \theta \\=(-3) \times \int_{0}^{2 \pi} [r^4/4]_{0}^{2} d \theta \\=-3 \int_0^{2 \pi} (4) d \theta \\=-3 \times 4 [ \theta]_0^{2 \pi} \\= -24 \pi$$
