Answer
$-6 \pi$
Work Step by Step
Applying Stoke's Theorem, we have
$\oint F \cdot dr=\iint _S (\nabla \times F) \cdot n d\sigma$
or, $\oint F \cdot dr=\int_{0}^{2 \pi} \int_{0}^{2} (-3x^2y^2) dy dx$
or, $\int_{0}^{2 \pi} \int_{0}^{2}-6 \sin^2 t+18 \cos^3 t dt=\int_{0}^{2 \pi} -6 \sin^2 t+18 \cos t-18 \cos t\sin^2 t dt$
This implies that
$[-3t+3 \sin t \cos t+18 \sin t-6 \sin^3 t]_{0}^{2 \pi}=-6 \pi-0=-6 \pi$
