Answer:

(i) For a parametric curve, the arc length is

\(s=\int_{0}^{2} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Differentiate the coordinates:

\(\frac{dx}{dt}=e^t-4\), \(\frac{dy}{dt}=4e^{t/2}\).

So

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(e^t-4)^2+(4e^{t/2})^2\).

Expanding gives

\(e^{2t}-8e^t+16+16e^t=e^{2t}+8e^t+16=(e^t+4)^2\).

Since \(e^t+4\gt 0\) on \(0\le t\le 2\), the speed is \(e^t+4\). Hence

\(s=\int_0^2 (e^t+4)\,dt=[e^t+4t]_0^2=(e^2+8)-(1)=e^2+7\).

(ii) The surface area formed by rotating about the \(x\)-axis is

\(S=2\pi\int_0^2 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Using \(y=8e^{t/2}\) and the speed found above,

\(S=2\pi\int_0^2 8e^{t/2}(e^t+4)\,dt\).

So

\(S=16\pi\int_0^2 \left(e^{3t/2}+4e^{t/2}\right)dt\).

Integrating term by term:

\(\int e^{3t/2}dt=\frac{2}{3}e^{3t/2}\), and \(\int 4e^{t/2}dt=8e^{t/2}\).

Therefore

\(S=16\pi\left[\frac{2}{3}e^{3t/2}+8e^{t/2}\right]_0^2\).

Substituting the limits gives

\(S=16\pi\left(\frac{2}{3}e^3+8e-\frac{2}{3}-8\right)=16\pi\left(\frac{2}{3}e^3+8e-\frac{26}{3}\right)\).