Answer
$\dfrac{195 \pi}{2}$
Work Step by Step
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$
Conversion to polar coordinates: $x=r \cos \theta; y= r \sin \theta$
We set up the line integral and find out the integrand of the double integral as follows:
$\oint_C [-3y^2+3x^2] dA=\iint_{D}(\dfrac{\partial (3x^2)}{\partial x}-\dfrac{\partial (-3y^2) }{\partial y})dA$
or, $=3 \int_{0}^{2 \pi} \int_{2}^{3} r^2 \times r dr d \theta$
or, $=3 \int_{0}^{2 \pi} \int_{2}^{3} r^3 dr d \theta$
or, $=3 \int_{0}^{2 \pi} [r^4/4]_{2}^{3} d \theta$
or, $=-3 \int_0^{2 \pi} (\dfrac{r^4}{4})_2^3 d \theta$
or, $=\dfrac{195 \pi}{2}$
