Ответ: S=3π/8 кв. ед.

Объяснение:

[tex]\displaystyle\\\left \{ {{x=cos^3t} \atop {y=sin^3t}} \right. \ \ \ \ \ \ t\in[0;2\pi]\\\\x'=(cos^3t)'=3*cos^2t*(cost)'=3*cos^2t*(-sint).\\\\y'=(sin^3t)'=3*sin^2t*(sint)'=3*sin^2t*cost.\ \ \ \ \ \ \Rightarrow\\\\[/tex]

[tex]\displaystyle\\S=\frac{1}{2}\int_0^{2\pi}(x*y'-x'*y)dt=\\\\=\frac{1}{2}\int_0^{2\pi}(cos^3t*3sin^2tcost-3cos^2t*(-sint)*sin^3t)dt=\\\\=\frac{3}{2}\int_0^{2\pi}(cos^4t*sin^2t+sin^4tcos^2t)dt=\frac{3}{2}\int_0^{2\pi}(sin^2t*cos^2t*(cos^2t+sin^2t)dt=\\\\[/tex]

[tex]\displaystyle\\=\frac{3}{2}\int_0^{2\pi}((sint*cost)^2*1)dt=\frac{3}{2}\int_0^{2\pi}(\frac{1}{2}sin2t)^2dt= \frac{3}{2}\int_0^{2\pi}\frac{1}{4} sin^22tdt=\\\\=\frac{3}{8}\int_0^{2\pi}sin^22tdt=\frac{3}{8}\int_0^{2\pi}\frac{1}{2} (1-cos4t)dt=\frac{3}{16}\int_0^{2\pi}(1-cos4t)dt=\\\\=\frac{3}{16}*(t-\frac{1}{4} sint)\ |^{2\pi}_0 =\frac{3}{16} *((2\pi-0)-\frac{1}{4}*(sin4*2\pi-sin4*0))=\\\\=\frac{3}{16}*( 2\pi-\frac{1}{4}* 0)=\frac{3}{16}*2\pi= \frac{3\pi }{8} .[/tex]