#### Answer

$\pi a^2 $

#### Work Step by Step

We have $r(t)=a \cos t i+a \sin t j$
Now, $x= a \cos t \implies dx=-a \sin t dt$
and $y= a \sin t \implies dy=a \cos t dt$
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{a^2 cos^2 t}{2} dt +\dfrac{a^2 sin^2 t}{2} dt =\int_{0}^{2 \pi} \dfrac{a^2}{2} dt=\dfrac{a^2}{2} \times (2 \pi) =\pi a^2 $