Answer
$\pi ab$
Work Step by Step
Consider $r(t)=a \cos t i+b \sin t j$
Now, $x= a \cos t $ and $y= a \sin t$
and $dx=-a \sin t dt ; \\ dy=a \cos t dt$
Now, $Area=\oint_{C} \dfrac{x dy}{2} - \dfrac{y dx}{2}$
$x= a \cos t $ and $y= a \sin t$
and $dx=-a \sin t dt ; \\ dy=a \cos t dt$
So, $Area=\int_{0}^{2 \pi} \dfrac{ab cos^2 t}{2} dt +\dfrac{ab sin^2 t}{2} dt $
or, $=\int_{0}^{2 \pi} \dfrac{ab}{2} dt$
and $\dfrac{ab}{2} \times (2 \pi) =\pi ab$
