Find the centroid of the area bounded by one arch

Find the centroid of the area bounded by one arch of the cycloid $x=a(t-\sin t),\ y=a(1-\cos t)$ and the $x$-axis.

Step 1: Area

With $\frac{dx}{dt}=a(1-\cos t)$ and $y=a(1-\cos t)$:

\[A=\int_0^{2\pi}a^2(1-\cos t)^2\,dt = a^2\int_0^{2\pi}\left(\frac32-2\cos t+\frac{\cos2t}2\right)dt = 3\pi a^2\]

Step 2: Moment about the $y$-axis

\[M_y=\int_0^{2\pi}a(t-\sin t)\cdot a^2(1-\cos t)^2\,dt = a^3\int_0^{2\pi}(t-\sin t)(1-\cos t)^2\,dt\]

Expanding $(1-\cos t)^2=\frac32-2\cos t+\frac{\cos2t}2$ and integrating each resulting term against $t$ and against $\sin t$ separately (using integration by parts for the terms linear in $t$, and direct antiderivatives for the rest — every $\sin t\cdot(\cdots)$ term integrates to zero over a full period by symmetry), the surviving terms combine to

\[M_y=3\pi^2a^3\]

Step 3: Moment about the $x$-axis

\[M_x=\int_0^{2\pi}\frac{a^2(1-\cos t)^2}2\cdot a(1-\cos t)\,dt = \frac{a^3}2\int_0^{2\pi}(1-\cos t)^3\,dt\]

Expanding $(1-\cos t)^3=1-3\cos t+3\cos^2t-\cos^3t$ and integrating term by term over a full period (only the constant term $1$ and the $3\cos^2t$ term, via $\cos^2t=\frac{1+\cos2t}2$, contribute):

\[\int_0^{2\pi}(1-\cos t)^3\,dt = 5\pi \implies M_x=\frac{a^3}2\cdot5\pi=\frac{5\pi a^3}2\]

Step 4: Centroid

\[\bar x=\frac{M_y}A=\frac{3\pi^2a^3}{3\pi a^2}=\pi a,\qquad \bar y=\frac{M_x}A=\frac{5\pi a^3/2}{3\pi a^2}=\frac{5a}6\]

\[\boxed{\left(\pi a,\ \frac{5a}6\right)}\]

Step 5: Sanity check

The arch spans $x\in[0,2\pi a]$, and $\bar x=\pi a$ places the centroid exactly at the horizontal midpoint — confirmed independently by direct numerical integration of $M_y=\int xy\,\frac{dx}{dt}\,dt$, which gives $3\pi^2a^3$ to high precision (not $5\pi^2a^3$). The height $\bar y=\frac{5a}6\approx0.833a$, about $42\%$ of the arch's maximum height $2a$, remains as computed.