Answer: The arc length is \(\sqrt2(e^{\pi}-e^{-\pi})\), or \(2\sqrt2\sinh\pi\).

The surface area is \(\displaystyle \frac{2\sqrt2\pi}{5}(e^{2\pi}-e^{-2\pi})\), or \(\displaystyle \frac{4\sqrt2\pi}{5}\sinh2\pi\).

Differentiate the parametric equations:

\(\dot x=2e^{2t}\cos2t-2e^{2t}\sin2t,\qquad \dot y=2e^{2t}\sin2t+2e^{2t}\cos2t.\)

Then

\(\dot x^2+\dot y^2=8e^{4t}.\)

So

\(\frac{ds}{dt}=2\sqrt2 e^{2t}.\)

The arc length is therefore

\(s=\int_{-\pi/2}^{\pi/2}2\sqrt2 e^{2t}\,dt=\sqrt2\left[e^{2t}\right]_{-\pi/2}^{\pi/2}=\sqrt2(e^{\pi}-e^{-\pi})=2\sqrt2\sinh\pi.\)

For the surface area about the \(x\)-axis, use

\(S=2\pi\int y\,ds.\)

Thus

\(S=2\pi\int_{-\pi/2}^{\pi/2}e^{2t}\sin2t\cdot 2\sqrt2e^{2t}\,dt=4\sqrt2\pi I,\)

where

\(I=\int_{-\pi/2}^{\pi/2}e^{4t}\sin2t\,dt.\)

Integrating by parts twice gives

\(I=\frac{e^{2\pi}-e^{-2\pi}}{10}.\)

Therefore

\(S=4\sqrt2\pi\cdot\frac{e^{2\pi}-e^{-2\pi}}{10}=\frac{2\sqrt2\pi}{5}(e^{2\pi}-e^{-2\pi})=\frac{4\sqrt2\pi}{5}\sinh2\pi.\)