Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 25

$4$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $S=\int_{0}^{\pi} \sqrt{(-\sin t)^2+(1+\cos t)^2} dt=\int_{0}^{\pi} \sqrt{2+2 \cos t} dt$ or, $\int_{0}^{\pi} \sqrt{2+2 (2\cos^2 (t/2)-1)} dt=\int_{0}^{\pi} 2\cos (t/2) dt$ Thus, $S=[4 \sin (t/2)]_{0}^{\pi} =4$