Answer: (i) The arc length is \(\frac{4}{3}\).

(ii) The surface area generated is \(\frac{11\pi}{9}\).

For a parametric curve, the arc length from \(t=0\) to \(t=1\) is

\(s=\int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Here \(x=t^2\) and \(y=t-\frac{1}{3}t^3\), so

\(\frac{dx}{dt}=2t\), and \(\frac{dy}{dt}=1-t^2\).

Then

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=4t^2+(1-t^2)^2=4t^2+1-2t^2+t^4=1+2t^2+t^4=(1+t^2)^2\).

Since \(0\le t\le 1\), \(\sqrt{(1+t^2)^2}=1+t^2\). Hence

\(s=\int_0^1 (1+t^2)\,dt=\left[t+\frac{t^3}{3}\right]_0^1=1+\frac13=\frac43\).

For the surface area when the curve is rotated about the \(x\)-axis, use

\(S=2\pi\int_0^1 y\,ds=2\pi\int_0^1 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

So

\(S=2\pi\int_0^1 \left(t-\frac13 t^3\right)(1+t^2)\,dt\).

Expanding the integrand gives

\(\left(t-\frac13 t^3\right)(1+t^2)=t+t^3-\frac13 t^3-\frac13 t^5=t+\frac23 t^3-\frac13 t^5\).

Therefore

\(S=2\pi\int_0^1 \left(t+\frac23 t^3-\frac13 t^5\right)dt\)

\(=2\pi\left[\frac12 t^2+\frac16 t^4-\frac1{18} t^6\right]_0^1\)

\(=2\pi\left(\frac12+\frac16-\frac1{18}\right)=2\pi\left(\frac{11}{18}\right)=\frac{11\pi}{9}.\)