Помогите, пожалуйста, вычислить площадь поверхности, образованной вращением кривой
y = sin x, 0 <= x <= pi вокруг оси Ох.

y = sin x, 0 <= x <= pi.
Используем формулу:
S = 2 * pi * int (0 pi) |y| * (1 + (y')^2)^(1/2) dx
y' = (sin x)' = cos x
Тогда
S = 2 * pi * int (0 pi) |sin x| * (1 + cos^2 x)^(1/2) dx =
= 2 * pi * int (0 pi) sin x * (1 + cos^2 x)^(1/2) dx =
= 2 * pi * int (0 pi) (1 + cos^2 x)^(1/2) d(-cos x) =
= -2 * pi * int (0 pi) (1 + cos^2 x)^(1/2) d(cos x) = | cos x = t | =
= -2 * pi * int (1 -1) (1 + t^2)^(1/2) dt = 2 * pi * int (-1 1) (1 + t^2)^(1/2) dt =
= 2 * pi * (t/2 * (1 + t^2)^(1/2) + 1/2 * ln |t + (1 + t^2)^(1/2)|)_{-1}^{1} =
= 2 * pi * ((1/2 * (1 + 1^2)^(1/2) + 1/2 * ln |1 + (1 + 1^2)^(1/2)|) -
- (-1/2 * (1 + (-1)^2)^(1/2) + 1/2 * ln |-1 + (1 + (-1)^2)^(1/2)|)) =
= 2 * pi * ((1/2 * 2^(1/2) + 1/2 * ln |1 + 2^(1/2)|) -
- (-1/2 * 2^(1/2) + 1/2 * ln |-1 + 2^(1/2)|)) =
= 2 * pi * (1/2 * 2^(1/2) + 1/2 * ln |1 + 2^(1/2)| +
+ 1/2 * 2^(1/2) - 1/2 * ln |-1 + 2^(1/2)|) =
= 2 * pi * (2^(1/2) + 1/2 * ln |(1 + 2^(1/2))/(-1 + 2^(1/2))|) =
= 2 * pi * (2^(1/2) + 1/2 * ln |(1 + 2^(1/2))^2|) =
= 2 * pi * 2^(1/2) + 2 * pi * 1/2 * 2 * ln (1 + 2^(1/2)) =
= 2 * 2^(1/2) * pi + 2 * pi * ln (1 + 2^(1/2))
Ответ: S = 2 * 2^(1/2) * pi + 2 * pi * ln (1 + 2^(1/2)).

Спасибо огромное