(i) To find the volume of revolution, we use the formula \(V = \pi \int (y + 1) \, dy\) from \(y = 0\) to \(y = h\).

Integrating, we have:

\(V = \pi \left[ \frac{y^2}{2} + y \right]_0^h\)

\(= \pi \left( \frac{h^2}{2} + h \right)\)

Thus, \(V = \pi \left( \frac{1}{2}h^2 + h \right)\).

(ii) To find the area of the shaded region when \(h = 3\), we calculate:

\(\int (y + 1)^{1/2} \, dy\) from \(y = 0\) to \(y = 3\).

Alternatively, \(6 - \int (x^2 - 1) \, dx\) from \(x = 1\) to \(x = 2\).

Integrating, we have:

\(\left[ \frac{2}{3}(y + 1)^{3/2} \right]_0^3\)

\(= \frac{2}{3} \left[ 8 - 1 \right]\)

\(= \frac{14}{3}\)