(i) The curve is given by \(y = \sqrt{9 - 2x^2}\). The derivative is \(\frac{dy}{dx} = \frac{1}{2\sqrt{9 - 2x^2}} \times (-4x)\).

At \(P(2, 1)\), \(x = 2\), so \(m = -4\). The gradient of the normal is \(\frac{1}{4}\).

The equation of the normal at \(P\) is \(y - 1 = \frac{1}{4}(x - 2)\).

Solving for \(A\) and \(B\):

\(A(-2, 0)\) and \(B(0, \frac{1}{2})\).

The midpoint of \(AP\) is \((0, \frac{1}{2})\), which is \(B\).

(ii) The volume of the solid of revolution is given by:

\(\int x^2 \, dy = \int \left( \frac{9}{2} - \frac{y^2}{2} \right) \, dy\).

\(= \frac{9y}{2} - \frac{y^3}{6}\).

Using limits from 1 to 3:

Volume = \(\left[ \frac{9y}{2} - \frac{y^3}{6} \right]_1^3 = \frac{44}{3} \pi\).