perimeter of astroid

The astroid

$$\sqrt[3]{x^2}+\sqrt[3]{y^2}=\sqrt[3]{a^2}$$

can be presented in the parametric form

$$x=a{\cos }^3\phi ,y=a{\sin }^3\phi ,$$

where the polar angle $\phi$ gets the values from $0$ to $2\pi$. The curve consists of four congruent arcs, one of which is obtained letting  $0\leqq \phi \leqq \frac{\pi }{2}$. Thus the whole perimeter of the astroid is

$$s=4{\int }_0^{\frac{\pi }{2}}\sqrt{{\left(\frac{dx}{d\phi }\right)}^2+{\left(\frac{dy}{d\phi }\right)}^2}𝑑\phi .$$

This expression is easily simplified to

$$s=12a{\int }_0^{\frac{\pi }{2}}\sin \phi \cos \phi d\phi$$

giving the result

$$s=6a{\int }_0^{\frac{\pi }{2}}\sin 2\phi d\phi =-3a\underset{0}{\overset{\frac{\pi }{2}}{/}}\cos 2\phi =6a.$$