Example 3. Find the mass and coordinates of the center of gravity of the arc of the astroid $x=10 \cos ^{3} t, y=10 \sin ^{3} t, 0 \leqslant t \leqslant \frac{\pi}{2}$, if its density $\rho=1$.
Solution. First, we determine the element of arc length $d l=\sqrt{x^{\prime 2}+y^{\prime 2}} d t . \quad$ We have $\quad x^{\prime}=-30 \cos ^{2} t \sin t, \quad y^{\prime}=30 \sin ^{2} t \cos t$, $d l=30 \sqrt{\cos ^{4} t \sin ^{2} t+\sin ^{4} t \cos ^{2} t} d t=30 \sin t \cos t d t=15 \sin 2 t d t$.

Next, we sequentially calculate the mass and static moments.

$$
\begin{gathered}
m=\int_{0}^{\pi / 2} \rho d l=15 \int_{0}^{\pi / 2} \sin 2 t d t=-\left.\frac{15}{2} \cos 2 t\right|_{0} ^{\pi / 2}=15 \\
M_{x}=\int_{0}^{\pi / 2} \rho x d l=10 \cdot 15 \int_{0}^{\pi / 2} \cos ^{3} t \cdot \sin 2 t=300 \int_{0}^{\pi / 2} \cos ^{4} t \sin t d t= \\
-\left.\frac{300}{5} \cos ^{5} t\right|_{0} ^{\pi / 2}=60 \\
M_{y}=\int_{0}^{\pi / 2} \rho y d l=300 \int_{0}^{\pi / 2} \sin ^{4} t \cos t d t=60 \\
x_{c}=\frac{M_{y}}{m}=\frac{60}{15}=4, y_{c}=\frac{M_{x}}{m}=4
\end{gathered}
$$

Answer. $m=15, M_{c}(4,4)$.