Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 658: 33

$\dfrac{52 \pi}{3}$

Here, we have $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt=\sqrt{1+(t+\sqrt 2)^2}=\sqrt{t^2+(2\sqrt 2)t+3}$ or, $S=\int_{-\sqrt 2}^{\sqrt 2} (2 \pi) x ds$ or, $S=\int_{-\sqrt 2}^{\sqrt 2} (2 \pi) (t+\sqrt 2) (\sqrt{t^2+(2\sqrt 2)t+3}) dt$ Plug $\sqrt{t^2+(2\sqrt 2) t+3}=k \implies (2t+2\sqrt 2) dt=dk$ $S=( \pi)\int_{1}^{9} \sqrt{k} dk=\dfrac{2\pi}{3}[k^{3/2}]_{1}^{9}$ Thus, $S=\dfrac{52 \pi}{3}$