Answer
$$2 \pi \ln 2+\frac{15 \pi}{8}$$
Work Step by Step
Since
$$\sqrt{1+y'^2}=\sqrt{ 1+ \sinh^2 x} =\cosh x$$
Then
\begin{aligned}
S &=2 \pi \int_{a}^{b} f(x)\sqrt{ 1+ f'^2(x)} d x\\
&=2 \pi \int_{-\ln 2}^{\ln 2} \cosh ^{2} x d x\\
&=\pi \int_{-\ln 2}^{\ln 2}(1+\cosh 2 x) d x\\
&=\left.\pi\left(x+\frac{1}{2} \sinh 2 x\right)\right|_{-\ln 2} ^{\ln 2} \\
&=\pi\left(\ln 2+\frac{1}{2} \sinh (2 \ln 2)+\ln 2-\frac{1}{2} \sinh (-2 \ln 2)\right)\\
&=2 \pi \ln 2+\pi \sinh (2 \ln 2)\\
&=2 \pi \ln 2+ \pi \frac{e^{2\ln 2}-e^{-2\ln 2} }{2}\\
&= 2 \pi \ln 2+\frac{15 \pi}{8}
\end{aligned}
