833. Find the center of gravity of an isosceles right triangle if the surface density at each of its points is proportional to the distance from it to the hypotenuse.
Solution. Let in a right isosceles triangle $ABC$ the hypotenuse $AB=2a$. Then, relative to the coordinate system depicted in Fig. 169, the equations of the legs $AC$ and $BC$ will be $y=x+a$ and $y=a-x$.

According to the problem's condition, the density $\delta=ky$ at the point $(x, y)$ of the triangle.

Next, using formulas (2), we will calculate the values of $m_{x}, m_{y}$, and $m$ for this triangle:

$$
\begin{aligned}
& m_{x}=\iint_{ABC} k y^{2} d x d y=k \int_{0}^{a} y^{2} d y \int_{y-a}^{a-y} d x=2 k \int_{0}^{a} y^{2}(a-y) d y= \\
&=\left.2 k\left(a \frac{y^{3}}{3}-\frac{y^{4}}{4}\right)\right|_{0} ^{a}=\frac{k a^{4}}{6} \\
& m_{y}=\iint_{ABC} k x y d x d y=k \int_{0}^{a} y d y \int_{-(a-y)}^{a-y} x d x=k \int_{0}^{a} y d y \cdot 0=0 \\
& m=\iint_{ABC} k y d x d y=k \int_{0}^{a} y d y \int_{y-a}^{a-y} d x=2 k \int_{0}^{a} y(a-y) d y=\frac{k a^{3}}{3}
\end{aligned}
$$

Therefore, $x_{c}=\frac{m_{y}}{m}=0, y_{c}=\frac{m_{x}}{m}=\frac{a}{2}$.

If the mass were distributed uniformly over the entire area of the triangle, its center of gravity would be located at the point of intersection of the medians $\left(0, \frac{a}{3}\right)$.