1323. Determine the force of water pressure on a vertical wall having the shape of a semicircle with radius $R=6$ m, the diameter of which is at the water surface.
Solution. The pressure force of a liquid on an area $S$ at a depth of immersion $x$ is equal to

$$
P=\gamma x S
$$

where $\gamma$ is the specific weight of the liquid.

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-258.jpg?height=277&width=420&top_left_y=427&top_left_x=78)

Fig. 90

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-258.jpg?height=219&width=499&top_left_y=485&top_left_x=600)

Fig. 91

We divide the semicircle into elementary strips by lines parallel to the water surface. We approximate the shaded strip as a rectangle with base $AB$ and height $dx$ (Fig. 91) and assume that all points of this strip are at the same depth $x (0 \leqslant x \leqslant R)$. This assumption is valid because $dx$ is small. The elementary area of the strip is

$$
d S=AB dx=2 \sqrt{OB^2-OC^2} dx=2 \sqrt{R^2-x^2} dx
$$

The pressure of the water on the strip at depth $x$ is equal to

$$
d P=\gamma x d S=2 \sqrt{R^2-x^2} \gamma x dx
$$

from which

$$
\begin{aligned}
P= & \int_{0}^{R} 2 \gamma x \sqrt{R^2-x^2} dx=2 \gamma \int_{0}^{R} x \sqrt{R^2-x^2} dx= \\
& =-2 \gamma \int_{R}^{0} t^2 dt=2 \gamma \int_{0}^{R} t^2 dt=\left.\frac{2 \gamma t^3}{3}\right|_{0} ^{R}=\frac{2 \gamma R^3}{3}
\end{aligned}
$$

(we applied the substitution $\sqrt{R^2-x^2}=t, \quad R^2-x^2=t^2, \quad x dx=$ $=-t dt$; if $x=0$, then $t=R$, if $x=R$, then $t=0$). It is known that the specific weight of water (weight of 1 cm $^3$) is $1 \kappa \Gamma$, therefore, the weight of 1 m $^3$ of water is $1000 \kappa \Gamma$. In the obtained result, $R$ is expressed in meters ($R=6$ m), so we need to set $\gamma=1000 \kappa \Gamma$. Taking this into account, we get

$$
\begin{aligned}
& P=\frac{2 \cdot 6^3 \cdot 1000}{3} \kappa \Gamma=144000 \kappa \Gamma \approx 144000 \cdot 9.81 \mu= \\
& =1412640 \mu \approx 14.13 \text{ bar }
\end{aligned}
$$

(newton - the unit of force (weight) in the International System of Units (SI); $1 \kappa \Gamma \approx 9.81 \kappa ; 1$ bar $=0.987$ atm).