The curve C has parametric equations x = ⅓t³ – ln t, y = ⅔t³, for 1 ≤ t ≤ 3.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2012). The mark or mark type is shown in brackets.

1. [B1] Differentiates: ẋ = t² - 1/t, ẏ = 2t²
2. [M1A1] Squares and adds: (ds/dt)² = (t² - 1/t)² + 4t⁴ = (t² + 1/t)² (Note: This simplification is mathematically incorrect based on the previous step and QP's y. The MS solution directly proceeds with √(t²+1/t)². To reconcile, one would need dy/dt=2, implying y=2t+C, not y=⅔t³.)
3. [M1] Uses arc length formula: s = ∫ (from 1 to 3) √(t² + 1/t)² dt = ∫ (from 1 to 3) (t² + 1/t) dt
4. [A1] Integrates: [t³/3 + ln t] (from 1 to 3)
5. [A1] Obtains result: (27/3 + ln 3) - (1/3 + ln 1) = 9 + ln 3 - 1/3 = 26/3 + ln 3 (= 9.77)
6. [M1] Uses surface area formula: S = 2π ∫ (from 1 to 3) y (ds/dt) dt = 2π ∫ (from 1 to 3) (7/3)t^(1/2) (t² + 1/t) dt (Note: This uses y = (7/3)t^(1/2) and ds/dt = t²+1/t, both are inconsistent with the QP's y and derived ds/dt from QP's x and y.)
7. [A1] Integrates: 2π ∫ (from 1 to 3) (7/3)(t^(5/2) + t^(-1/2)) dt = 8π/3 [2/7 t^(7/2) + 2/3 t^(3/2)] (from 1 to 3)
8. [M1] Inserts limits: 8π/3 [(2/7 * 3^(7/2) + 2/3 * 3^(3/2)) - (2/7 * 1^(7/2) + 2/3 * 1^(3/2))]
9. [A1] Obtains result: π(160√3/3 - 64/27) = (283 or = 90.0π)