Example 2.70. Calculate the coordinates of the center of gravity of a plane figure located in the first quadrant and bounded by the lines $y=a x^{3}$ and $y=a$, if the density is constant, $\rho=1$.
Solution. It is easy to obtain the coordinates of the intersection point of the lines $y=a x^{3}$ and $y=a: x=1, y=a$. According to formulas (2.46), (2.47), we get:

$$
\begin{gathered}
M_{x}=0.5 \int_{0}^{1}\left(a^{2}-a^{2} x^{6}\right) d x=0.5\left(a^{2} x-\frac{a^{2} x^{7}}{7}\right)\bigg|_{0} ^{1}=\frac{3 a^{2}}{7} \\
M_{y}=\int_{0}^{1} x\left(a-a x^{3}\right) d x=\left(\frac{a x^{2}}{2}-\frac{a x^{5}}{5}\right)\bigg|_{0} ^{1}=0.3 a \\
S=\int_{0}^{1}\left(a-a x^{3}\right) d x=\left(a x-\frac{a x^{4}}{4}\right)\bigg|_{0} ^{1}=\frac{3 a}{4}
\end{gathered}
$$

From this, $x_{c}=\frac{M_{y}}{S}=0.4, y_{c}=\frac{M_{x}}{S}=\frac{4 a}{7}$.