Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 31

$8 \pi^2$

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