Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 25

$4$

Here, we have $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $S=\int_{0}^{\pi} \sqrt{(-\sin t)^2+(1+\cos t)^2} dt$ $S=\int_{0}^{\pi} \sqrt{2+2 \cos t} dt=\int_{0}^{\pi} \sqrt{2+2 [2\cos^2 (\dfrac{t}{2})-1]} dt$ This implies that $S=\int_{0}^{\pi} 2\cos (t/2) dt$ and $S=[4 \sin (\dfrac{t}{2})]_{0}^{\pi} =4$