{ x = 8 cos³ t,
{ y = 8 sin³ t, x = 1 (x ≥ 1).
8cos3t=1 ⇒ cost=1/2 ⇒ t1=–π/3; t2=π/3
dx=8·3cos2t·(cost)`dt=–24cos2t·sint dt
S= ∫ t2t1y(t)x`(t)dt=
= ∫π/3–π/38·sin3t·(–24cos2t)·sint dt=
=–192 ∫π/3–π/3 sin4t·cos2tdt=
понижаем степень:
=–192 ∫π/3–π/3[m](\frac{1-cos2t}{2})^2\cdot \frac{1+cos2t}{2}[/m]dt=
=–24 ∫π/3–π/3(1–cos22t–cos2t+cos32t)dt=
cos32t=cos22t·cos2t=(1–sin22t)·cos2t и раз понижаем четную степень:
=–24 ∫π/3–π/3(1–[m]\frac{1+cos4t}{2}[/m]–cos2t+cos2t–sin22t·cos2t)dt=
=–24 ∫π/3–π/3([m]\frac{1}{2}-\frac{cos4t}{2}[/m]–sin22t·cos2t)dt=
=–24·([m]\frac{1}{2}t-\frac{sin4t}{8}-\frac{1}{2}\cdot \frac{sin^32t}{3}[/m])|π/3–π/3=
пределы–то подставите? самостоятельно....