Example 3. Calculate the force of water pressure $P$ on a vertically submerged plate having the shape of a triangle $ABC$ with base $AC=b$ and height $BD=h$, assuming that the vertex $B$ of this triangle lies on the free surface of the liquid, and the base $AC$ is parallel to it (Fig. 5.23).

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-225.jpg?height=408&width=505&top_left_y=147&top_left_x=39)

Fig. 5.23
Solution. Let $MN$ be the width of the plate at the level $BE=x$. From the similarity of triangles $MBN$ and $ABC$, we find

$$
\frac{MN}{AC}=\frac{BE}{BD} \text { or } \frac{y}{b}=\frac{x}{h},
$$

from which

$$
y=\frac{bx}{h}
$$

Based on formula (5.25), we obtain

$$
P=\gamma \int_{0}^{h} x \frac{bx}{h} dx = \frac{\gamma b}{h} \int_{0}^{h} x^{2} dx = \left. \frac{\gamma b}{h} \cdot \frac{x^{3}}{3} \right|_{0}^{h} = \frac{\gamma b}{h} \cdot \frac{h^{3}}{3} = \frac{\gamma b h^{3}}{3}
$$

It is known that the weight of a cubic meter of water is 1000 kg. Therefore, if $b$ and $h$ are expressed in meters, then $\gamma=9,80665 \cdot 10^{3}$ N and $P=9,80665 \cdot 10^{3} \frac{b h^{2}}{3}$.

For example, if $b=6$ m and $h=2$ m, then $P=78453.2$ N.

## Problems

1. The velocity of a body thrown vertically upwards with an initial velocity $v_{0}$, neglecting air resistance, is determined by the formula

$$
v=v_{0}-gt
$$

where $t$ is the elapsed time; $g$ is the acceleration due to gravity. At what distance from the initial position will the body be after $t$ seconds of flight?

2. The velocity of a point is $v=te^{-0.01t} m/s$. Find the distance traveled by the point from the start of the motion until it comes to a complete stop.
3. A rocket projectile rises vertically upwards. Assuming that with constant thrust, the acceleration of the rocket increases according to the law $y=\frac{A}{a-bt}(a-bt>0)$, find the velocity at any moment of time $t$, if the initial velocity is zero. Also find the height reached by the rocket at the moment of time $t=t_{1}$.
4. Calculate the work required to pump water out of a vertical cylindrical tank with a base radius $R$ and height $H$.
5. What work is required to raise a body of mass $m$ from the surface of the Earth with radius $R$ to a height $h$?
6. Calculate the kinetic energy of a disk of mass $M$ and radius $R$ rotating with angular velocity $\omega$ about an axis passing through its center perpendicular to its plane.
7. What work is required to stop an iron sphere of radius $R=2 \mathcal{M}$, rotating with angular velocity $\boldsymbol{\omega}=\frac{100}{3} \pi$ rad/s around its diameter (specific weight of iron $\gamma=7,8 \cdot 10^{3} \kappa g / \mathbf{M}^{3}$)?
8. A vertical dam has the shape of a trapezoid. Calculate the force of water pressure on the dam if the upper base of the dam $a=80 m$, the lower base $b=50 m$, and the height of the dam $h=20 m$.
9. Find the force of liquid pressure, with specific weight $\gamma$, on a vertical ellipse with axes $2a$ and $2b$, the center of which is submerged in the liquid at a level $h$, and the major axis $2a$ of the ellipse is parallel to the liquid surface $(h \geq b)$.

Answers

1. $v_{0} t-\frac{g t^{2}}{2}$. 2. $s=10^{4} M$. 3. $v=\frac{A}{b} \ln \left(\frac{a}{a-b t}\right)$. $h=\frac{A}{b^{2}}\left[b t_{1}-(a-b t_{1}) \ln \frac{a}{a-b t_{1}}\right]$. 4. $A=\frac{\pi \gamma}{2} R^{2} H^{2}$. Hint. The elementary force (gravitational force) is equal to the weight of the water in the volume of a layer of thickness $d x$, i.e., $d F=\pi \gamma R^{2} d x$, where $\gamma$ is the weight of a unit volume of water. Therefore, the elementary work of the force is $d A=\gamma \pi R^{2}(H-x) d x$, where $x$ is the water level. 5. $A=\frac{m g h}{1+\frac{h}{2}}$. The force acting on a body of mass $m$ is $F=k \frac{m M}{r^{2}}$, where $r$ is the distance from the center of the Earth. Since at $r=R$ we have $F=m g$, then $k M=g R^{2}$. The required work is $A=\int_{R}^{R+h} k \frac{m M}{r^{2}} d r=k m M\left(\frac{1}{R}-\frac{1}{R+h}\right)=\frac{m g h}{1+\frac{h}{R}}$. 6. $\frac{1}{4} M R^{2} \omega^{2}$. Solution. The kinetic energy of a particle of the disk $d k=\frac{m v^{2}}{2}=\frac{\rho r^{2} \omega^{2}}{2} d \sigma$, where $\sigma$ is the element of area; $r$ is the distance from the axis of rotation; $\rho$ is the surface density, $\rho=\frac{M}{\pi R^{2}}$. Thus, $d k=\frac{M \omega^{2}}{2 \pi R^{2}} r^{2} d \sigma$. Therefore, $k=\frac{M \omega^{2}}{R^{2}} \int_{0}^{R} r^{3} d r=\frac{M R^{2} \omega^{2}}{4}$. 7. $k=\frac{M}{5} R^{2} \omega^{2}=22.5 \cdot 10^{8} \partial \mathscr{}$. Hint: The amount of work required is equal to the kinetic energy reserve. 8. $P=\frac{(a+2 b) h^{2}}{b} \approx 117679.8 \cdot 10^{3} \mathrm{N}$. 9. $p=a b \gamma \pi h$.

## § 5.6. Improper Integrals

Integrals with infinite limits and integrals of unbounded functions are called improper integrals.

Integrals with infinite limits. By definition, we have

$$
\int_{a}^{+\infty} f(x) d x=\lim _{b \rightarrow+\infty} \int_{a}^{b} f(x) d x=\lim _{b \rightarrow+\infty}[F(b)-F(a)]
$$

where

$$
F^{\prime}(x)=f(x)
$$

If the limit on the right-hand side of equation (5.26) exists and is finite, then the improper integral is called convergent; otherwise, it is called divergent.

If $|f(x)| \leq \varphi(x)$ and the integral $\int_{a}^{+\infty} \varphi(x) d x$ converges, then the integral (5.26) also converges (comparison test).

Similarly, we define:

$$
\int_{-\infty}^{b} f(x) d x=\lim _{a \rightarrow-\infty} \int_{a}^{b} f(x) d x
$$

$$
\int_{-\infty}^{+\infty} f(x) d x=\lim _{a \rightarrow-\infty} \int_{a}^{c} f(x) d x+\lim _{b \rightarrow \infty} \int_{c}^{b} f(x) d x
$$

Integrals of unbounded functions. If the function $f(x)$ is unbounded in any neighborhood of the point $c$ on the interval $[a, b]$ and is continuous for $a \leq x<c$ and $c<x \leq b$, then by definition we have

$$
\int_{a}^{b} f(x) d x=\lim _{\varepsilon \rightarrow 0} \int_{a}^{c-\varepsilon} f(x) d x+\lim _{\eta \rightarrow 0} \int_{c+\eta}^{b} f(x) d x
$$

where $\varepsilon$ and $\eta$ vary independently of each other. In the case $c=b$ or $c=a$, we get

$$
\int_{a}^{b} f(x) d x=\lim _{\varepsilon \rightarrow 0} \int_{a}^{b-\varepsilon} f(x) d x
$$

or

$$
\int_{a}^{b} f(x) d x=\lim _{\varepsilon \rightarrow 0} \int_{a+\varepsilon}^{b} f(x) d x
$$

The improper integral (5.29) is called convergent or divergent depending on whether the limits defining it exist or not.

If there exists a continuous function $F(x)$ on $[a, b]$ such that $F^{\prime}(x)=f(x)$ for $x \neq c$, then

$$
\int_{a}^{b} f(x) d x=F(b)-F(a)
$$