(i) Start with the equation \(y = \frac{2}{\sqrt{x+1}}\).

Square both sides: \(y^2 = \frac{4}{x+1}\).

Rearrange to find \(x\): \(x+1 = \frac{4}{y^2}\).

Thus, \(x = \frac{4}{y^2} - 1\).

(ii) Integrate \(\int \left( \frac{4}{y^2} - 1 \right) \, dy\).

The integral is \(\left[ -\frac{4}{y} - y \right]\).

Apply limits from 1 to 2: \(\left[ -\frac{4}{2} - 2 \right] - \left[ -\frac{4}{1} - 1 \right]\).

Calculate: \(\left[ -2 - 2 \right] - \left[ -4 - 1 \right] = -4 + 5 = 1\).

(iii) Volume of revolution: \(\pi \int x^2 \, dy = \pi \int \left( \frac{16}{y^4} - \frac{8}{y^2} + 1 \right) \, dy\).

The integral is \(\pi \left[ -\frac{16}{3y^3} + \frac{8}{y} + y \right]\).

Apply limits from 1 to 2: \(\pi \left[ -\frac{16}{24} + 4 + 2 \right] - \pi \left[ -\frac{16}{3} + 8 + 1 \right]\).

Calculate: \(\pi \left[ -\frac{2}{3} + 6 \right] - \pi \left[ -\frac{16}{3} + 9 \right] = \pi \left[ \frac{16}{3} \right] = \frac{5\pi}{3}\).