403. Determine the force of water pressure on a vertical parabolic segment, the base of which is $4 \mathrm{m}$ and is located on the water surface, while the vertex is at a depth of $4 \mathrm{m}$ (Fig. 204).
Solution. We have $|B A|=2 x=4$ (m). Point $A$ in the chosen coordinate system has coordinates $(2 ; 4)$. The equation of the parabola relative to this system is $y = a x^{2}$ or $4=a \cdot 2^{2}$, from which $a=1$, i.e., $y=x^{2}$.

Consider an elementary area $d S$ at a distance $y$ from the origin. The length of this area is $2 x$, and its area is $d S=2 x d y$, where $d y$ is the width of the area. This area will experience a pressure force

$$
d P=9.81 \gamma(4-y) d S=9.81 \gamma(4-y) 2 x d y
$$

Summing all the elementary forces of pressure as $y$ changes from 0 to 4 and taking the limit as $\Delta y \rightarrow 0$, the force of pressure on the entire segment will be expressed by the following integral:

$$
P=9.81 \gamma \int_{0}^{4}(4-y) 2 x d y
$$

To find this integral, express $x$ in terms of $y$; since $y=x^{2}$, then $x=\sqrt{y}$. Then we get

$$
\begin{aligned}
& P=19.62 \cdot 1000 \int_{0}^{4}(4-y) \sqrt{y} d y=19620\left(\int_{0}^{4} 4 \sqrt{y} d y-\int_{0}^{4} y \sqrt{y} d y\right)= \\
&=19620\left(4 \int_{0}^{4} y^{1 / 2} d y-\int_{0}^{4} y^{3 / 2} d y\right)=\left.19620\left(\frac{4 y^{3 / 2}}{3 / 2}-\frac{y^{5 / 2}}{5 / 2}\right)\right|_{0} ^{4}= \\
&=\left.19620\left(\frac{8}{3} y^{3 / 2}-\frac{2}{5} y^{5 / 2}\right)\right|_{0} ^{4}=19620\left(\frac{8 \cdot 2}{3}-\frac{2 \cdot 32}{5}\right)= \\
&=\frac{19620 \cdot 128}{15}=167424 \text { (N). }
\end{aligned}
$$