# perimeter of astroid

The astroid

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

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

 $s=6a\int_{0}^{\frac{\pi}{2}}\sin{2\varphi}\,d\varphi=-3a\!% \operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad\frac{\pi}{2}}\!\cos{2% \varphi}=6a.$
Title perimeter of astroid PerimeterOfAstroid 2013-03-22 17:14:06 2013-03-22 17:14:06 pahio (2872) pahio (2872) 7 pahio (2872) Example msc 26B15 perimetre of astroid Parameter ClairautsEquation SubstitutionNotation