832. Find the mass of a plate having the shape of an ellipse, if the surface density at each point of the plate is proportional to its distance $r$ from the minor axis of the ellipse and at $r=1$ it is equal to $\lambda$.
Solution. Let the semi-axes of the ellipse be denoted by $a$ and $b (a > b)$, and choose the axes $O x$ and $O y$ of the rectangular coordinate system to coincide with the axes of the ellipse. Then the equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

According to the problem, at the point $M(x, y)$ on the plate, the density $\delta(M)=\lambda|x|$.

By formula (1), the mass of the right half of the plate is

$$
\begin{gathered}
\frac{1}{2} m=\iint_{D} \lambda x d x d y=\lambda \int_{-b}^{b} d y \int_{\prime \prime}^{\frac{a}{b} \sqrt{b^{2}-y^{2}}} x d x=\frac{\lambda a^{2}}{2 b^{2}} \int_{-b}^{b}\left(b^{2}-y^{2}\right) d y= \\
=\left.\frac{\lambda a^{2}}{2 b^{2}}\left(b^{2} y-\frac{y^{3}}{3}\right)\right|_{-b} ^{b}=\frac{2}{3} a^{2} b \lambda .
\end{gathered}
$$

Therefore, the mass of the entire plate is $m=\frac{4}{3} a^{2} b \lambda$.