The astroid $$\sqrt[3]{x^2}+\sqrt[3]{y^2} = \sqrt[3]{a^2}$$ can be presented in the parametric form $$x = a\cos^3\varphi,\quad y = a\sin^3\varphi,$$ where the polar angle $\varphi$ gets the values from $0$ to $2\pi$ . The curve consists of four congruent arcs, one of which is obtained letting $0\leqq \varphi \leqq \frac{\pi}{2}$ . Thus the whole perimeter of the astroid is $$s = 4\int_0^{\frac{\pi}{2}}\!\sqrt{\left(\frac{dx}{d\varphi}\right)^2+\left(\frac{dy}{d\varphi}\right)^2}\,d\varphi.$$ This expression is easily simplified to $$s = 12a\int_0^{\frac{\pi}{2}}\sin\varphi\cos\varphi\,d\varphi$$ giving the result $$s = 6a\int_0^{\frac{\pi}{2}}\sin{2\varphi}\,d\varphi = -3a\!\sijoitus{0}{\quad \frac{\pi}{2}}\!\cos{2\varphi} = 6a.$$