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