Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 979: 26

$\pi ab$

We have $r(t)=a \cos t i+b \sin t j$ 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: $A=\int_{0}^{2 \pi} \dfrac{ab cos^2 t}{2} dt +\dfrac{ab sin^2 t}{2} dt =\int_{0}^{2 \pi} \dfrac{ab}{2} dt=\dfrac{ab}{2} \times (2 \pi) =\pi ab$