Example 2.69. Find the center of gravity of the arc of the circle $x^{2}+$ $+y^{2}=R^{2}$, located in the first quadrant, if the density $\rho=\rho_{0} x y$.
Solution. Differentiating the equation of the circle as an implicit function, we find

$$
\begin{gathered}
2 x+2 y y^{\prime}=0 \Rightarrow y^{\prime}=-\frac{x}{y}, \Rightarrow \\
\Rightarrow d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x=\sqrt{1+\left(-\frac{x}{y}\right)^{2}} d x= \\
=\frac{R}{y} d x=\frac{R}{\sqrt{R^{2}-x^{2}}} d x
\end{gathered}
$$

Using formulas (2.43) and (2.47)

$$
\begin{gathered}
M_{y}=\int_{0}^{R} \rho_{0} x \sqrt{R^{2}-x^{2}} \frac{x R}{\sqrt{R^{2}-x^{2}}} d x=\rho_{0} R \int_{0}^{R} x^{2} d x= \\
=\left.\rho_{0} R \frac{x^{3}}{3}\right|_{0} ^{R}=\rho_{0} \frac{R^{4}}{3} \\
M_{x}=\int_{0}^{R} \rho_{0} x \sqrt{R^{2}-x^{2}} \sqrt{R^{2}-x^{2}} \frac{R}{\sqrt{R^{2}-x^{2}}} d x= \\
=\rho_{0} R \int_{0}^{R} x \sqrt{R^{2}-x^{2}} d x=-0.5 \rho_{0} R \int_{0}^{R} \sqrt{R^{2}-x^{2}} d\left(R^{2}-x^{2}\right)= \\
=-\left.\rho_{0} \frac{R}{3}\left(R^{2}-x^{2}\right)^{3 / 2}\right|_{0} ^{R}=\rho_{0} \frac{R^{4}}{3} .
\end{gathered}
$$

The coordinates of the center of mass $x_{c}=y_{c}=\frac{2 R}{3}$.