655. Determine the water pressure on a vertical rectangular gate with a base of 18 m and a height of 6 m.
Solution. The magnitude $p$ of the liquid pressure on a horizontal platform depends on the depth of its submersion $x$, i.e., the distance from the platform to the surface of the liquid: $p=\delta a x$; $\delta$ - specific weight of the liquid, $a$ - area of the platform.

Guided by the general scheme (II) for applying the definite integral to the calculation of quantities, we divide the gate by

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-208.jpg?height=116&width=362&top_left_y=1333&top_left_x=824)

Fig. 117 at depth $x$ horizontally (Fig. 117). Then the pressure of the water on the upper part of the gate will be some function $p(x)$. Let's find the differential $(p$ of this function, i.e., the approximate value (main part) of its increment $\Delta p$ when the depth $x$ changes by a small amount $d x$.

Assume, due to the smallness of $d x$, that all points of the shaded strip are at depth $x$, i.e., that it is located at depth $x$ in a horizontal plane. Then the approximate value of the water pressure on this strip will be equal to the weight of the water column with this strip as its base and height - the depth $x$:

$\Delta p \approx d p=18 \delta x d x=18 x d x$. (Specific weight of water $\delta=1^{*}$.)

According to the problem, the depth $x$ varies over the interval $0 \leqslant x \leqslant 6$. Therefore, the sought pressure $P$ on the entire gate will be found by integrating $d p$ from 0 to 6:

$P=18 \int_{0}^{6} x d x=\left.9 x^{2}\right|_{0} ^{6}=324 T \approx 324000 \cdot 9.81 \mathrm{N} \approx 3178440 \mathrm{N}^{* *} \approx 3.18 \mathrm{MN}$.