Answer
$S$ = $\pi(\frac{15}{16}+\ln2)$
Work Step by Step
$\frac{dx}{dy}$ = $\frac{1}{2}$$(e^{y}-e^{-y})$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$2\pi(x)(\sqrt {1+(\frac{dx}{dy})^{2}}$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$2\pi(\frac{1}{2})(e^{y}+e^{-y})(\sqrt {1+(\frac{1}{2}(e^{y}-e^{-y}))^{2}}$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$\pi(e^{y}+e^{-y})(\sqrt {1+\frac{1}{4}e^{2y}-\frac{1}{2}+\frac{1}{4}e^{-2y}}$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$\pi(e^{y}+e^{-y})(\sqrt {\frac{1}{4}e^{2y}+\frac{1}{2}+\frac{1}{4}e^{-2y}}$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$\pi(e^{y}+e^{-y})(\sqrt {(\frac{1}{2}(e^{y}+e^{-y}))^{2}}$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$\pi(e^{y}+e^{-y})\frac{1}{2}(e^{y}+e^{-y})$$dy$
$S$ = $\int_{{\,0}}^{{\,\ln2}}$$\frac{\pi}{2}(e^{2y}+2+e^{-2y})$$dy$
$S$ = $\frac{\pi}{2}(\frac{1}{2}e^{2y}+2y-\frac{1}{2}e^{-2y})$$|_{{\,0}}^{{\,\ln2}}$
$S$ = $\frac{\pi}{2}[(2+2\ln2-\frac{1}{8})-(\frac{1}{2}+0-\frac{1}{2})]$
$S$ = $\pi(\frac{15}{16}+\ln2)$
