Question

Ребят, очень нужна помощь!!!!!! Даю почти 25 балов

Answer 1

$r=a\, sin^3\frac{\phi }{3}\\\\r\geq 0\; \; \to \; \; \; a\, sin^3\frac{\phi }{3}\geq 0\; ,\; \; sin\frac{\phi }{3}\geq 0\; ,\; \; 0\leq \frac{\phi }{3}\leq \pi \; ,\; \; 0\leq \phi \leq 3\pi \; .\\\\r'=3a\, sin^2\frac{\phi }{3}\cdot cos\frac{\phi }{3}\cdot \frac{1}{3}=a\, sin^2\frac{\phi }{3}\cdot cos\frac{\phi }{3}\\\\r^2+(r')^2=a^2\, sin^6\frac{\phi }{3}+a^2\, sin^4\frac{\phi }{3}\cdot cos^2\frac{\phi }{3}=a^2\cdot sin^4\frac{\phi }{3}\cdot (\underbrace {sin^2\frac{\phi }{3}+cos^2\frac{\phi}{3}}_{1})=\\\\=a^2\, sin^4\frac{\phi }{3}$

$L=\int\limits^{\phi _2}_{\phi _1}\, \sqrt{r^2+(r')^2}\; d\phi =\int\limits^{3\pi }_0\, \sqrt{a^2\, sin^4\frac{\phi }{3}}\; d\phi =a\int\limits^{3\pi }_0\, sin^2\frac{\phi }{3}\; d\phi =\\\\=a\int\limits^{3\phi }_0\, \frac{1-cos\frac{2\phi }{3}}{2}\; d\phi =\frac{a}{2}\int\limits^{3\phi }_0\, (1-cos\frac{2\phi }{3})\; d\phi =\frac{a}{2}\cdot (\phi - \frac{3}{2}\, sin\frac{2\phi }{3})\Big |_0^{3\pi }=\\\\=\frac{a}{2}\cdot (3\pi -\frac{3}{2}\, sin2\pi )=\frac{3\pi a}{2}$