## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{52 \pi}{3}$
Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $dS=\sqrt{(1)^2+(t+\sqrt 2)^2}=\sqrt{t^2+2\sqrt 2t+3}$ or, $S=\int_{-\sqrt 2}^{\sqrt 2} (2 \pi) x ds=\int_{-\sqrt 2}^{\sqrt 2} (2 \pi) (t+\sqrt 2) (\sqrt{t^2+2\sqrt 2t+3}) dt$ Suppose $p=\sqrt{t^2+2\sqrt 2t+3} \implies dp=(2t+2\sqrt 2) dt$ $S=( \pi)\int_{1}^{9} \sqrt{p} dp=[\dfrac{2\pi u^{3/2}}{3}]_{1}^{9}$ Thus, $S=\dfrac{52 \pi}{3}$