Find the centroid of the homogeneous figure bounded
Find the centroid of the homogeneous figure bounded by one arch of the cycloid $x=a(t-\sin t),\,y=a(1-\cos t),\,0\le t\le2\pi$, and the $x$-axis.
Explanation of Method
The arch is symmetric about the vertical line through its midpoint, fixing $\bar x$ immediately; the area and static moments are then computed using the parametric area formula $dA=y\,\dfrac{dx}{dt}\,dt$.
- Use symmetry to get $\bar x=\pi a$.
- Compute the area under the arch.
- Compute the static moment $M_x$.
- Verify $\bar x$ directly via $M_y$.
Step 1 · Symmetry
The arch is symmetric about $x=\pi a$ (at $t=\pi$), so $\bar x=\pi a$.
Step 2 · Area
With $\dfrac{dx}{dt}=a(1-\cos t)$:
$$ A = \int_0^{2\pi}y\,\frac{dx}{dt}\,dt = a^2\int_0^{2\pi}(1-\cos t)^2\,dt = a^2\int_0^{2\pi}\left(\frac32-2\cos t+\frac{\cos2t}2\right)dt = 3\pi a^2. $$
Step 3 · Static moment $M_x$
$$ M_x = \frac12\int_0^{2\pi}y^2\,\frac{dx}{dt}\,dt = \frac{a^3}2\int_0^{2\pi}(1-\cos t)^3\,dt = \frac{a^3}2\left(2\pi+3\pi\right) = \frac{5\pi a^3}2, $$
using $\int_0^{2\pi}(1-\cos t)^3\,dt = 2\pi-3\cdot0+3\pi-0=5\pi$. Hence $\bar y=\dfrac{M_x}A=\dfrac{5\pi a^3/2}{3\pi a^2}=\dfrac{5a}6$.
Step 4 · Verification via $M_y$
$$ M_y = \int_0^{2\pi}xy\,\frac{dx}{dt}\,dt = a^3\int_0^{2\pi}(t-\sin t)(1-\cos t)^2\,dt = a^3(3\pi^2-0) = 3\pi^2a^3, $$
so $\bar x=\dfrac{M_y}A=\dfrac{3\pi^2a^3}{3\pi a^2}=\pi a$, matching Step 1.
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=1.3$): $A\approx15.9279$, $\bar x\approx4.08407$, $\bar y\approx1.08333$, matching $3\pi a^2,\,\pi a,\,\dfrac{5a}6$ exactly.
Result
The centroid of the region under one arch of the cycloid is: