Answer
$\dfrac{195 \pi}{2}$
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:
$\oint_C [-3y^2+3x^2] dA=\iint_{D}(\dfrac{\partial (3x^2)}{\partial x}-\dfrac{\partial (-3y^2) }{\partial y})dA$
$= \int_{0}^{2 \pi} \int_{2}^{3} (r^2) (r) \ dr \ d \theta \times 3$
$= \int_{0}^{2 \pi} \int_{2}^{3} r^3 \ dr \ d \theta \times 3$
$= \int_{0}^{2 \pi} [\dfrac{r^4}{4}]_{2}^{3} d \theta $
$=(-3) \int_0^{2 \pi} (\dfrac{r^4}{4})_2^3 d \theta$
$=\dfrac{195 \pi}{2}$
