Answer: (i) The arc length is \(\frac{72}{5}\).
(ii) The surface area generated is \(190\pi\).
For a parametric curve, the arc length from \(t=1\) to \(t=4\) is \(\int_1^4 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).
We differentiate:
\(x=\frac{2}{5}t^{5/2}-2t^{1/2}\) so \(\frac{dx}{dt}=t^{3/2}-t^{-1/2}\).
\(y=\frac{4}{3}t^{3/2}\) so \(\frac{dy}{dt}=2t^{1/2}\).
Hence
\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(t^{3/2}-t^{-1/2})^2+4t\).
Expanding the square gives
\(t^3-2t+t^{-1}+4t=t^3+2t+t^{-1}=(t^{3/2}+t^{-1/2})^2\).
Since \(t\gt 0\) on \(1\le t\le 4\), we take
\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=t^{3/2}+t^{-1/2}\).
Therefore
\(s=\int_1^4 (t^{3/2}+t^{-1/2})\,dt = \left[\frac{2}{5}t^{5/2}+2t^{1/2}\right]_1^4\).
So
\(s=\left(\frac{2}{5}\cdot 4^{5/2}+2\cdot 4^{1/2}\right)-\left(\frac{2}{5}\cdot 1^{5/2}+2\cdot 1^{1/2}\right)\)
\(=\left(\frac{64}{5}+4\right)-\left(\frac{2}{5}+2\right)=\frac{72}{5}\).
For the surface area when the curve is rotated about the \(x\)-axis, use
\(S=2\pi\int_1^4 y\,ds=2\pi\int_1^4 y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).
So
\(S=2\pi\int_1^4 \frac{4}{3}t^{3/2}(t^{3/2}+t^{-1/2})\,dt\)
\(=\frac{8}{3}\pi\int_1^4 (t^3+t)\,dt\).
Integrating gives
\(S=\frac{8}{3}\pi\left[\frac14 t^4+\frac12 t^2\right]_1^4\).
Thus
\(S=\frac{8}{3}\pi\left(\left(64+8\right)-\left(\frac14+\frac12\right)\right)=190\pi\).
