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