682. Find the center of gravity of a homogeneous figure (plate) bounded by the parabola $\sqrt{x}+\sqrt{y}=\sqrt{a}$ and the coordinate axes.
Solution. The given homogeneous figure is symmetric with respect to the bisector of the first coordinate angle (Fig. 127), therefore $x_{c}=y_{c}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-224.jpg?height=169&width=518&top_left_y=342&top_left_x=146)

Fig. 126

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-224.jpg?height=323&width=334&top_left_y=191&top_left_x=764)

Fig. 127

Let's compute the integrals contained in the first 113 formulas (2):

$$
\begin{aligned}
& I_{1}=\int_{a}^{b} x y d x=\int_{0}^{a} x\left(a^{\frac{1}{2}}-x^{\frac{1}{2}}\right)^{2} d x=\int_{0}^{a}\left(a x-2 a^{\frac{1}{2}} x^{\frac{3}{2}}+x^{2}\right) d x= \\
& =\left.\left(\frac{1}{2} a x^{2}-\frac{4}{5} a^{\frac{1}{2}} x^{\frac{b}{2}}+\frac{1}{3} x^{3}\right)\right|_{0} ^{a}=\frac{a^{3}}{30} \\
& I_{2}=\int_{a}^{b} y d x=\int_{0}^{a}\left(a^{\frac{1}{2}}-x^{\frac{1}{2}}\right)^{2} d x=\int_{0}^{a}\left(a-2 a^{-\frac{1}{2}} x^{\frac{1}{2}}+x\right) d x=\frac{a^{2}}{6}
\end{aligned}
$$

Therefore, $x_{C}=y_{C}=\frac{l_{1}}{I_{2}}=\frac{a}{5}$.