Problem 137
Easy Physical Applications

Find the static moment with respect to the coordinate

Find the static moment with respect to the coordinate axes of the arc of the astroid $x^{2/3}+y^{2/3}=a^{2/3}$ in the first quadrant.

Explanation of Method

Use the parametrization $x=a\cos^3t,\ y=a\sin^3t$ with arc-length element $3a\cos t\sin t\,dt$, and integrate $y\,ds$ and $x\,ds$ over $t\in[0,\pi/2]$.

Step 1: Set up

$x=a\cos^3t,\ y=a\sin^3t$, $ds=3a\cos t\sin t\,dt$, $t\in[0,\pi/2]$.

Step 2: Moment about the $x$-axis

\[M_x=\int_0^{\pi/2}a\sin^3t\cdot3a\cos t\sin t\,dt = 3a^2\int_0^{\pi/2}\sin^4t\cos t\,dt = 3a^2\left[\frac{\sin^5t}5\right]_0^{\pi/2}=\frac{3a^2}5\]

Step 3: Moment about the $y$-axis

By the $x\leftrightarrow y$ symmetry of the first-quadrant astroid arc:

\[M_y=\int_0^{\pi/2}a\cos^3t\cdot3a\cos t\sin t\,dt = 3a^2\int_0^{\pi/2}\cos^4t\sin t\,dt = \frac{3a^2}5\]

Step 4: Arc length and centroid

\[L=\int_0^{\pi/2}3a\cos t\sin t\,dt = \frac{3a}2\int_0^{\pi/2}\sin2t\,dt=\frac{3a}2\]

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

\[\boxed{M_x=M_y=\frac{3a^2}5,\qquad\text{centroid}=\left(\frac{2a}5,\frac{2a}5\right)}\]


Result

Final answer:

\[\boxed{M_x = M_y = \frac{3a^2}{5}}\]