392. A rectangular plate is immersed in water, positioned vertically. Its horizontal side is 1 m, the vertical side is $2 \mathrm{m}$. The top side is at a depth of 0.5 m. Determine the force of water pressure on the plate.
Solution. Here $y=1, \quad a=0.5, \quad b=2+0.5=2.5$ (m), $\gamma=$ $=1000 \mathrm{kr} / \mathrm{m}^{3}$. Therefore,

$$
\begin{gathered}
P=9810 \int_{a}^{b} x y d x=9810 \int_{a}^{b} x d x=\left.9810 \frac{x^{2}}{2}\right|_{0.5} ^{2.5}=9810 \frac{2.5^{2}-0.5^{2}}{2}= \\
=29430 \text { (N) }
\end{gathered}
$$