Find the centroid of the arc of the astroid

Find the centroid of the arc of the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ in the first quadrant.

Step 1: Recall the results from Problem 137

\[L=\frac{3a}2,\qquad M_x=M_y=\frac{3a^2}5\]

Step 2: Compute the centroid

\[\bar x=\frac{M_y}L=\frac{3a^2/5}{3a/2}=\frac{2a}5,\qquad \bar y=\frac{M_x}L=\frac{2a}5\]

\[\boxed{\text{Centroid: }\left(\frac{2a}5,\frac{2a}5\right)}\]

Step 3: Interpretation

By the astroid's symmetry about the line $x=y$, the centroid necessarily lies on that line. It sits at $\frac{2a}5=0.4a$ along each axis, noticeably closer to the origin than the midpoint $\left(\frac a2,\frac a2\right)$ of the chord joining the arc's endpoints $(a,0)$ and $(0,a)$ — a consequence of the astroid's concave, inward-curving shape, which pulls more of the arc length toward the origin.