Answer
$-24 \pi$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
The integrand of the double integral becomes:
$\oint_C [-3x^2-3y^2] dA=\iint_{D}(\dfrac{\partial (-3x^2)}{\partial x}-\dfrac{\partial (3y^2) }{\partial y})dA$
and $\oint_C [-3x^2-3y^2] dA=-3 \times \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times r dr d \theta\\=-3 \int_{0}^{2 \pi} [\dfrac{r^4}{4}]_{0}^{2} d \theta\\=-3\times \int_0^{2 \pi} (4) d \theta \\=-3 \times 4 [\theta]_0^{2 \pi} \\= -24 \pi$
