(a) [3 marks]

(i) Solve \( y = x^2 + 2 \implies x = \sqrt{y – 2} \), \( y \) from 2 to 4.
Area \( R = \int_2^4 \sqrt{y – 2} \, dy \) (M1, A1).
(ii) Integrate: \( \int \sqrt{y – 2} \, dy = \frac{2}{3} (y – 2)^{3/2} \), evaluate from 2 to 4.
\( \left[ \frac{2}{3} (y – 2)^{3/2} \right]_2^4 = \frac{2}{3} (4 – 2)^{3/2} – \frac{2}{3} (2 – 2)^{3/2} = \frac{2}{3} \cdot 2^{3/2} \approx 1.886 \) (A1).

(b) [3 marks]

Volume by rotation: \( V = \pi \int_2^4 (y – 2) \, dy \) (M1).
Integrate: \( \int (y – 2) \, dy = \frac{y^2}{2} – 2y \), evaluate from 2 to 4.
\( \pi \left[ \frac{y^2}{2} – 2y \right]_2^4 = \pi \left( \frac{16}{2} – 8 – \frac{4}{2} + 4 \right) = \pi (8 – 4) = 4\pi \).
Corrected: \( V = \pi \int_2^4 (\sqrt{y – 2})^2 \, dy = \pi \int_2^4 (y – 2) \, dy = \pi \left[ \frac{y^2}{2} – 2y \right]_2^4 = 2\pi \) (A1, A1).