Answer
$\dfrac{3 \pi}{8}$
Work Step by Step
We have $ dx=-3 \cos^2 t \sin t dt $ and $dy=3 \sin^2 t \cos t dt$
Consider, $A=\oint_{C} \dfrac{x}{2} dy - \dfrac{y}{2} dx$
After substituting the variables $ dx=-3 \cos^2 t \sin t dt $ and $dy=3 \sin^2 t \cos t dt$ , we have:
$Area=\int_{0}^{2 \pi} \dfrac{1}{2} \times (\cos^3 t) \times (3 \sin^2 t \cos t dt) +\dfrac{sin^3 t}{2} ( -3 \cos^2 t \sin t dt) $
or, $=\dfrac{3}{2} \times \int_{0}^{2 \pi} \cos^2 t \times \sin^2 t (\cos^2 t+\sin^2 t) dt$
or, $=\dfrac{3}{16} \times \int_0^{2 \pi} [1-\cos 4t] dt$
or, $= [\dfrac{3t}{16}-\dfrac{3 \sin 4t}{64}]_0^{2 \pi}$
or, $ A=\dfrac{3 \pi}{8}$
