Find the arc length of the astroid x^(2/3) + y^(2/3) =

Find the arc length of the astroid $x^{2/3} + y^{2/3} = a^{2/3}$.

We parametrize the astroid and exploit its four-fold symmetry.

Step 1: Parametrize

\[x = a\cos^3t, \qquad y = a\sin^3t\]

Step 2: Compute derivatives

$\frac{dx}{dt}=-3a\cos^2t\sin t$, $\frac{dy}{dt}=3a\sin^2t\cos t$.

Step 3: Simplify the integrand

\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = 9a^2\cos^2t\sin^2t(\cos^2t+\sin^2t) = 9a^2\cos^2t\sin^2t\]

For $t\in\left[0,\frac\pi2\right]$, both $\sin t,\cos t\ge0$, so the square root is $3a\sin t\cos t$.

Step 4: Integrate over one quadrant and multiply by 4

\[L = 4\int_0^{\pi/2}3a\sin t\cos t\,dt = 12a\int_0^{\pi/2}\sin t\cos t\,dt\]

With $u=\sin t$: $\int_0^{\pi/2}\sin t\cos t\,dt = \left[\frac{u^2}2\right]_0^1=\frac12$.

\[L = 12a\cdot\frac12 = 6a\]

\[\boxed{L = 6a}\]

Summary

\[\boxed{L = 6a}\]

Key insight: The astroid's four cusps (at the coordinate axes) don't prevent the total arc length from being finite and remarkably simple — exactly $6a$, or $\frac34$ of the perimeter ($8a$) of the square in which the astroid is inscribed.