680. Find the center of gravity of a quarter circle $x^{2}+y^{2}=a^{2}$, located in the first quadrant, if the linear density at each of its points is proportional to the product of the coordinates of the point.
Solution. From the equation of the circle, we find $y^{\prime}$, then $d l$:

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

Next, we compute the integrals contained in formulas (1), assuming, according to the condition, $\delta=k x y$:

$$
\begin{gathered}
\int_{(A)}^{(B)} \delta x d l==\int_{0}^{a} k x y \cdot x \cdot \frac{a}{y} d x=k a \int_{0}^{a} x^{2} d x=\left.\frac{k a}{3} x^{3}\right|_{0} ^{a}=\frac{k a^{4}}{3} \\
\int_{(A)}^{(B)} \delta y d l=k a \int_{0}^{a} x y d x=k a \int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x= \\
=\frac{k a}{2} \int_{a}^{0}\left(a^{2}-x^{2}\right)^{\frac{1}{2}} d\left(a^{2}-x^{2}\right)=\left.\frac{k a}{3}\left(a^{2}-x^{2}\right)^{\frac{3}{2}}\right|_{a} ^{0}=\frac{k a^{4}}{3} \\
\int_{(A)}^{(B)} \delta d l=k a \int_{0}^{a} x d x=\left.\frac{k a}{2} x^{2}\right|_{0} ^{a}=\frac{k a^{3}}{2}
\end{gathered}
$$

Substituting the values of the integrals into formulas (1), we get

$$
x_{c}=y_{c}=\frac{2}{3} a \text {. }
$$

Obviously, the found point does not lie on the given arc, but is located below it.