Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 979: 25

$\pi a^2 $

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 $