Answer
$\dfrac{3 \pi}{8}$
Work Step by Step
We have:
$r(t)=xi+y j=\cos^3 t i+\sin^3 t j$
Now, $ dx=-3 \cos^2 t \sin t dt $ and $dy=3 \sin^2 t \cos t dt$
Thus, the area becomes:
$A=\oint_{C} \dfrac{x}{2} dy - \dfrac{y}{2} dx$
After substituting the values, we have:
$A=\int_{0}^{2 \pi} \dfrac{\cos^3 t}{2} (3 \sin^2 t \cos t dt) +\dfrac{\sin^3 t}{2} ( -3 \cos^2 t \sin t dt) \\=\dfrac{3}{2} \int_{0}^{2 \pi} \cos^2 t \sin^2 t (\cos^2 t+\sin^2 t) dt \\=\dfrac{3}{8} \int_0^{2 \pi} \sin^2 (2t) dt \\=\dfrac{3}{16} \int_0^{2 \pi} [1-\cos 4t] dt=\dfrac{3}{16} [t-\dfrac{\sin 4t}{4}]_0^{2 \pi} \\=\dfrac{3 \pi}{8}$
