681. Find the center of gravity of a homogeneous cycloid arch

$$
x=a(t-\sin t), \quad y=a(1-\cos t), \quad \text{fig. } 126
$$
Solution. The given homogeneous arc is symmetric with respect to the line $x=\pi a$. Therefore, the center of gravity of the arc lies on this line, i.e., $x_{c}=\pi a$. To determine $y_{c}$, we find the differential of the arc of the cycloid

$$
d l=\sqrt{x^{2}+y^{2}} d t=\sqrt{a^{2}(1-\cos t)^{2}+a^{2} \sin ^{2} t} d t=2 a \sin \frac{t}{2} d t
$$

and compute the integrals contained in the second of formulas (1):

$$
\begin{gathered}
I_{1}=\int_{(A)}^{(B)} \delta y d l=2 \delta a^{2} \int_{0}^{2 \pi}(1-\cos t) \sin \frac{t}{2} d t= \\
=\left.2 \delta a^{2}\left(\int \sin \frac{t}{2} d t-\int \cos t \sin \frac{t}{2} d t\right)\right|_{0} ^{2 \pi}= \\
\left.=2 \delta a^{2}\left\{\left.2 \int \sin \frac{t}{2} d \frac{t}{2}-\frac{1}{2} \int \right\rvert\, \sin \frac{3}{2} t+\sin \left(-\frac{t}{2}\right)\right] d t\right\}\left.\right|_{0} ^{2 \pi}= \\
=\left.2 \delta a^{2}\left(-3 \cos \frac{t}{2}+\frac{1}{3} \cos \frac{3}{2} t\right)\right|_{0} ^{2 \pi}=\frac{32}{3} \delta a^{2} . \\
I_{2}=\int_{(A)}^{(B)} \delta d l=2 \delta a \int_{0}^{2 \pi} \sin \frac{t}{2} d t \cdots-\left.4 \varepsilon a \cos \frac{t}{2}\right|_{0} ^{2 \pi}=8 \delta a .
\end{gathered}
$$

By formula (1), $y_{C}=-\frac{1}{3} a$.