Помогите, пожалуйста, вычислить площадь фигуры, ограниченной кривой
x = 3 * (t - sin t), y = 3 * (1 - cos t), pi <= t <= 2 * pi

x = 3 * (t - sin t), y = 3 * (1 - cos t), pi <= t <= 2 * pi.
S = int (pi 2 * pi) x * y' dt = int (pi 2 * pi) 3 * (t - sin t) * (3 * (1 - cos t))' dt =
= int (pi 2 * pi) 3 * (t - sin t) * 3 * (1 - cos t)' dt =
= 9 * int (pi 2 * pi) (t - sin t) * sin t dt =
= 9 * int (pi 2 * pi) t * sin t dt - 9 * int (pi 2 * pi) sin^2 t dt =
= 9 * int (pi 2 * pi) t d(-cos t) - 9 * int (pi 2 * pi) (1 - cos 2t)/2 dt =
= -9 * int (pi 2 * pi) t d(cos t) - 9/2 * int (pi 2 * pi) (1 - cos 2t) dt =
= -9 * (t * cos t)_{pi}^{2 * pi} + 9 * int (pi 2 * pi) cos t dt -
- 9/2 * (t - 1/2 * sin 2t)_{pi}^{2 * pi} =
= -9 * (2 * pi * cos (2pi) - pi * cos pi) + 9 * (sin t)_{pi}^{2 * pi} -
- 9/2 * ((2 * pi - 1/2 * sin (4pi)) - (pi - 1/2 * sin (2pi))) =
= -9 * (2 * pi + pi) + 9 * (sin (2pi) - sin pi) - 9/2 * (2 * pi - pi) =
= -9 * 3 * pi - 9/2 * pi = -27 * pi - 9/2 * pi = -63/2 * pi.
Ответ: S = 63/2 * pi.

Спасибо!