Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 31

$8 \pi^2$

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ This implies that $dS=\int_{0}^{2\pi} \sqrt{(-\sin t)^2+( \cos t)^2}=1$ Now, $S=\int_{0}^{2\pi} (2 \pi)(2+\sin t)(1) dt$ Thus, $S=(2 \pi)[(2t-\cos t)]_{0}^{2\pi} =(2 \pi)[(4 \pi-(1-1)]=8 \pi^2$