Physical applications of the definite integral work of a variable force. Mechanical applications of the definite integral. Surface area of rotation

The definite integral (OI) is widely used in practical applications of mathematics and physics.

In particular, in geometry, with the help of ROI, the areas of simple figures and complex surfaces, the volumes of bodies of revolution and bodies of arbitrary shape, the lengths of curves in the plane and in space are found.

In physics and theoretical mechanics, RI is used to calculate the static moments, masses and centers of mass of material curves and surfaces, to calculate the work of a variable force along a curved path, etc.

Area of a flat figure

Let some plane figure in the Cartesian rectangular coordinate system $xOy$ be bounded from above by the curve $y=y_(1) \left(x\right)$, from below by the curve $y=y_(2) \left(x\right)$ , and on the left and right by vertical lines $x=a$ and $x=b$ respectively. In general, the area of such a figure is expressed using the OR $S=\int \limits _(a)^(b)\left(y_(1) \left(x\right)-y_(2) \left(x\right )\right)\cdot dx $.

If some flat figure in the Cartesian rectangular coordinate system $xOy$ is bounded on the right by the curve $x=x_(1) \left(y\right)$, on the left - by the curve $x=x_(2) \left(y\right) $, and below and above by horizontal lines $y=c$ and $y=d$, respectively, then the area of such a figure is expressed using the OI $S=\int \limits _(c)^(d)\left(x_(1 ) \left(y\right)-x_(2) \left(y\right)\right)\cdot dy $.

Let a plane figure (a curvilinear sector) considered in a polar coordinate system be formed by the graph of a continuous function $\rho =\rho \left(\phi \right)$, as well as by two rays passing at angles $\phi =\alpha $ and $\phi =\beta $ respectively. The formula for calculating the area of such a curvilinear sector is: $S=\frac(1)(2) \cdot \int \limits _(\alpha )^(\beta )\rho ^(2) \left(\phi \right )\cdot d\phi$.

Curve arc length

If on the segment $\left[\alpha ,\; \beta \right]$ curve is given by the equation $\rho =\rho \left(\phi \right)$ in polar coordinates, then the length of its arc is calculated using the OR $L=\int \limits _(\alpha )^ (\beta )\sqrt(\rho ^(2) \left(\phi \right)+\rho "^(2) \left(\phi \right)) \cdot d\phi $.

If the curve on the segment $\left$ is given by the equation $y=y\left(x\right)$, then the length of its arc is calculated using the OR $L=\int \limits _(a)^(b)\sqrt(1 +y"^(2) \left(x\right)) \cdot dx $.

If on the segment $\left[\alpha ,\; \beta \right]$ the curve is given parametrically, i.e. $x=x\left(t\right)$, $y=y\left(t\right)$, then the length of its arc is calculated using the OR $L=\ int \limits _(\alpha )^(\beta )\sqrt(x"^(2) \left(t\right)+y"^(2) \left(t\right)) \cdot dt $.

Calculation of the body volume from the areas of parallel sections

Let it be necessary to find the volume of a spatial body whose point coordinates satisfy the conditions $a\le x\le b$, and for which the cross-sectional areas $S\left(x\right)$ by planes perpendicular to the $Ox$ axis are known.

The formula for calculating the volume of such a body is $V=\int \limits _(a)^(b)S\left(x\right)\cdot dx $.

Volume of a body of revolution

Let a non-negative continuous function $y=y\left(x\right)$ be given on the segment $\left$, forming a curvilinear trapezoid (KrT). If we rotate this CRT around the $Ox$ axis, then a body is formed, called the body of revolution.

Calculation of the volume of a body of revolution is a special case of calculating the volume of a body from the known areas of its parallel sections. The corresponding formula is $V=\int \limits _(a)^(b)S\left(x\right)\cdot dx =\pi \cdot \int \limits _(a)^(b)y^( 2) \left(x\right)\cdot dx$.

Let some plane figure in the Cartesian rectangular coordinate system $xOy$ be bounded from above by the curve $y=y_(1) \left(x\right)$, from below by the curve $y=y_(2) \left(x\right)$ , where $y_(1) \left(x\right)$ and $y_(2) \left(x\right)$ are non-negative continuous functions, and vertical lines $x=a$ and $x= b$ respectively. Then the volume of the body formed by the rotation of this figure around the $Ox$ axis is expressed by the OR $V=\pi \cdot \int \limits _(a)^(b)\left(y_(1)^(2) \left(x \right)-y_(2)^(2) \left(x\right)\right)\cdot dx $.

Let some plane figure in the Cartesian rectangular coordinate system $xOy$ be bounded on the right by the curve $x=x_(1) \left(y\right)$, on the left - by the curve $x=x_(2) \left(y\right)$ , where $x_(1) \left(y\right)$ and $x_(2) \left(y\right)$ are non-negative continuous functions, and horizontal lines $y=c$ and $y= d$ respectively. Then the volume of the body formed by the rotation of this figure around the $Oy$ axis is expressed by the OI $V=\pi \cdot \int \limits _(c)^(d)\left(x_(1)^(2) \left(y \right)-x_(2)^(2) \left(y\right)\right)\cdot dy $.

Surface area of a body of revolution

Let a non-negative function $y=y\left(x\right)$ with a continuous derivative $y"\left(x\right)$ be given on the segment $\left$. This function forms a KrT. If we rotate this KrT around the axis $Ox $, then it itself forms a body of revolution, and the arc KrT is its surface.The surface area of such a body of revolution is expressed by the formula $Q=2\cdot \pi \cdot \int \limits _(a)^(b)y\left( x\right)\cdot \sqrt(1+y"^(2) \left(x\right)) \cdot dx $.

Suppose that the curve $x=\phi \left(y\right)$, where $\phi \left(y\right)$ is a non-negative function defined on the segment $c\le y\le d$, is rotated around the axis $Oy$. In this case, the surface area of the formed body of revolution is expressed as OR $Q=2\cdot \pi \cdot \int \limits _(c)^(d)\phi \left(y\right)\cdot \sqrt(1+\phi "^(2) \left(y\right)) \cdot dy $.

Physical applications of OI

To calculate the distance traveled at time $t=T$ with a variable speed $v=v\left(t\right)$ of a material point that started moving at time $t=t_(0) $, use the OR $S =\int \limits _(t_(0) )^(T)v\left(t\right)\cdot dt $.
To calculate the work of a variable force $F=F\left(x\right)$ applied to a material point moving along a rectilinear path along the $Ox$ axis from the point $x=a$ to the point $x=b$ (direction of the force coincides with the direction of travel) use the ROI $A=\int \limits _(a)^(b)F\left(x\right)\cdot dx $.
The static moments about the coordinate axes of the material curve $y=y\left(x\right)$ on the interval $\left$ are expressed by the formulas $M_(x) =\rho \cdot \int \limits _(a)^(b)y \left(x\right)\cdot \sqrt(1+y"^(2) \left(x\right)) \cdot dx $ and $M_(y) =\rho \cdot \int \limits _(a )^(b)x\cdot \sqrt(1+y"^(2) \left(x\right)) \cdot dx $, where the linear density $\rho $ of this curve is assumed to be constant.
The center of mass of a material curve is a point at which its entire mass is conditionally concentrated in such a way that the static moments of the point relative to the coordinate axes are equal to the corresponding static moments of the entire curve as a whole.

The formulas for calculating the coordinates of the center of mass of a plane curve are $x_(C) =\frac(\int \limits _(a)^(b)x\cdot \sqrt(1+y"^(2) \left(x\ right)) \cdot dx )(\int \limits _(a)^(b)\sqrt(1+y"^(2) \left(x\right)) \cdot dx ) $ and $y_(C) =\frac(\int \limits _(a)^(b)y\left(x\right)\cdot \sqrt(1+y"^(2) \left(x\right)) \cdot dx )( \int \limits _(a)^(b)\sqrt(1+y"^(2) \left(x\right)) \cdot dx ) $.

The static moments of a material flat figure in the form of KrT relative to the coordinate axes are expressed by the formulas $M_(x) =\frac(1)(2) \cdot \rho \cdot \int \limits _(a)^(b)y^(2) \left(x\right)\cdot dx $ and $M_(y) =\rho \cdot \int \limits _(a)^(b)x\cdot y\left(x\right)\cdot dx $.
The coordinates of the center of mass of a material flat figure in the form of KrT, formed by the curve $y=y\left(x\right)$ on the interval $\left$, are calculated by the formulas $x_(C) =\frac(\int \limits _(a )^(b)x\cdot y\left(x\right)\cdot dx )(\int \limits _(a)^(b)y\left(x\right)\cdot dx ) $ and $y_( C) =\frac(\frac(1)(2) \cdot \int \limits _(a)^(b)y^(2) \left(x\right)\cdot dx )(\int \limits _ (a)^(b)y\left(x\right)\cdot dx ) $.

Topic 6.10. Geometric and physical applications of the definite integral

1. The area of a curvilinear trapezoid bounded by a curve y \u003d f (x) (f (x)> 0), straight lines x \u003d a, x \u003d b and a segment [a, b] of the Ox axis, is calculated by the formula

2. The area of the figure bounded by the curves y = f (x) and y = g (x) (f (x)< g (x)) и прямыми х= a , x = b , находится по формуле

3. If the curve is given by the parametric equations x \u003d x (t), y \u003d y (t), then the area of \u200b\u200bthe curvilinear trapezoid bounded by this curve and the straight lines x \u003d a, x \u003d b, is found by the formula

4. Let S (x) be the cross-sectional area of the body by a plane perpendicular to the Ox axis, then the volume of the part of the body enclosed between the planes x \u003d a and x \u003d perpendicular to the axis is found by the formula

5. Let a curvilinear trapezoid bounded by a curve y \u003d f (x) and straight lines y \u003d 0, x \u003d a and x \u003d b, rotates around the Ox axis, then the volume of the body of revolution is calculated by the formula

6. Let a curvilinear trapezoid bounded by a curve x \u003d g (y) and

straight lines x \u003d 0, y \u003d c and y \u003d d, rotates around the axis О y, then the volume of the body of revolution is calculated by the formula

7. If a plane curve is related to a rectangular coordinate system and is given by the equation y \u003d f (x) (or x \u003d F (y)), then the length of the arc is determined by the formula

1. The area of a flat figure.

Area of a curvilinear trapezoid bounded by a non-negative function f(x), the abscissa axis and straight lines x = a, x = b, is defined as S = ∫ a b f x d x .

Area of a curvilinear trapezoid

The area of a figure bounded by a function f(x) intersecting the x-axis is defined by the formula S = ∑ i: f x ≥ 0 ∫ x i - 1 x i f x d x - ∑ i: f x< 0 ∫ x i - 1 x i | f x | d x , где x i are zeros of the function. In other words, to calculate the area of this figure, you need to split the segment function zeros f(x) into parts, integrate the function f for each of the resulting intervals of constancy, add separately the integrals over the segments on which the function f takes different signs, and subtract the second from the first.

2. Area of a curvilinear sector.

Area of a curved sector Consider a curve ρ = ρ (φ) in the polar coordinate system, where ρ (φ) is continuous and non-negative on [α; β] function. Figure bounded by a curve ρ (φ) and rays φ = α , φ = β , is called a curvilinear sector. The area of the curved sector is equal to S = 1 2 ∫ α β ρ 2 φ d φ .

3. Volume of a body of revolution.

Volume of a body of revolution

Let the body be formed by rotation around the axis OX of a curvilinear trapezoid bounded by a continuous one on the segment function f(x). Its volume is expressed by the formula V = π ∫ a b f 2 x d x .

On the problem of finding the volume of a body from the cross-sectional area

Let the body be enclosed between planes x = a And x = b, and the area of its section by a plane passing through the point x, is continuous on the segment function σ(x). Then its volume is V = ∫ a b σ x d x .

4. The length of the arc of the curve.

Let the curve r → t = x t , y t , z t be given. Then the length of its segment, bounded by the values t = α And t = β is expressed by the formula S = ∫ α β x ′ t 2 + y ′ t 2 + z ′ t 2 dt .

Arc length of a plane curve In particular, the length of a plane curve defined on the coordinate plane OXY equation y=f(x), a ≤ x ≤ b, is expressed by the formula S = ∫ a b 1 + f ′ x 2 dx .

5. Surface area of rotation.

Surface area of rotation Let the surface be defined by rotation about the OX axis of the graph of the function y=f(x), a ≤ x ≤ b, and function f has a continuous derivative on this segment. Then the area of the surface of revolution is determined by the formula Π = 2 π ∫ a b f x 1 + f ′ x 2 d x .

The area of a curvilinear trapezoid bounded from above by the graph of a function y=f(x), left and right - straight x=a And x=b respectively, from below - the axis Ox, is calculated by the formula

Area of a curvilinear trapezoid bounded on the right by the graph of a function x=φ(y), top and bottom - straight y=d And y=c respectively, on the left - the axis Oy:

The area of a curvilinear figure bounded from above by a graph of a function y 2 \u003d f 2 (x), below - graph of the function y 1 \u003d f 1 (x), left and right - straight x=a And x=b:

The area of a curvilinear figure bounded on the left and right by function graphs x 1 \u003d φ 1 (y) And x 2 \u003d φ 2 (y), top and bottom - straight y=d And y=c respectively:

Consider the case when the line limiting the curvilinear trapezoid from above is given by the parametric equations x = φ 1 (t), y \u003d φ 2 (t), Where α ≤ t ≤ β, φ 1 (α)=a, φ 1 (β)=b. These equations define some function y=f(x) on the segment [ a, b]. The area of a curvilinear trapezoid is calculated by the formula

Let's move on to a new variable x = φ 1 (t), Then dx = φ" 1 (t) dt, A y=f(x)=f(φ 1 (t))=φ 2 (t), hence \begin(displaymath)

Area in polar coordinates

Consider a curvilinear sector OAB, bounded by the line given by the equation ρ=ρ(φ) in polar coordinates, two beams OA And OB, for which φ=α , φ=β .

We divide the sector into elementary sectors OM k-1 M k ( k=1, …, n, M 0 =A, Mn=B). Denote by Δφk angle between beams OM k-1 And OM k forming angles with the polar axis φk-1 And φk respectively. Each of the elementary sectors OM k-1 M k replace with a circular sector with radius ρ k \u003d ρ (φ "k), Where φ" k- angle value φ from the interval [ φk-1 , φk], and the central angle Δφk. The area of the last sector is expressed by the formula .

expresses the area of the "stepped" sector, which approximately replaces the given sector OAB.

Sector area OAB is called the limit of the area of the "stepped" sector at n→∞ And λ=max Δφ k → 0:

Because , That

Curve arc length

Let on the segment [ a, b] a differentiable function is given y=f(x), whose graph is the arc . Line segment [ a,b] split into n parts dots x 1, x2, …, xn-1. These points will correspond to the points M1, M2, …, Mn-1 arcs, connect them with a broken line, which is called a broken line inscribed in an arc. The perimeter of this broken line is denoted by s n, that is

Definition. The length of the arc of the line is the limit of the perimeter of the polyline inscribed in it, when the number of links M k-1 M k increases indefinitely, and the length of the largest of them tends to zero:

where λ is the length of the largest link.

We will count the length of the arc from some of its points, for example, A. Let at the point M(x,y) arc length is s, and at the point M"(x+Δx,y+Δy) arc length is s+Δs, where, i>Δs - arc length. From a triangle MNM" find the length of the chord: .

From geometric considerations it follows that

that is, the infinitely small arc of the line and the chord that subtends it are equivalent.

Let's transform the formula expressing the length of the chord:

Passing to the limit in this equality, we obtain a formula for the derivative of the function s=s(x):

from which we find

This formula expresses the differential of the arc of a plane curve and has a simple geometric sense: expresses the Pythagorean theorem for an infinitesimal triangle MTN (ds=MT, ).

The differential of the arc of the space curve is given by

Consider an arc of a space line given by the parametric equations

Where α ≤ t ≤ β, φ i (t) (i=1, 2, 3) are differentiable functions of the argument t, That

Integrating this equality over the interval [ α, β ], we obtain a formula for calculating the length of this line arc

If the line lies in a plane Oxy, That z=0 for all t∈[α, β], That's why

In the case when the flat line is given by the equation y=f(x) (a≤x≤b), Where f(x) is a differentiable function, the last formula takes the form

Let the flat line be given by the equation ρ=ρ(φ) (α≤φ≤β ) in polar coordinates. In this case, we have the parametric equations of the line x=ρ(φ) cos φ, y=ρ(φ) sin φ, where the polar angle is taken as a parameter φ . Because the

then the formula expressing the length of the arc of the line ρ=ρ(φ) (α≤φ≤β ) in polar coordinates has the form

body volume

Let us find the volume of a body if the area of any cross section of this body perpendicular to a certain direction is known.

Let us divide this body into elementary layers by planes perpendicular to the axis Ox and defined by the equations x=const. For any fixed x∈ known area S=S(x) cross section of this body.

Elementary layer cut off by planes x=x k-1, x=x k (k=1, …, n, x 0 =a, xn=b), we replace it with a cylinder with a height ∆x k =x k -x k-1 and base area S(ξk), ξk ∈.

The volume of the specified elementary cylinder is expressed by the formula Δvk =E(ξk)Δxk. Let's sum up all such products

which is the integral sum for the given function S=S(x) on the segment [ a, b]. It expresses the volume of a stepped body, consisting of elementary cylinders and approximately replacing the given body.

The volume of a given body is the limit of the volume of the specified stepped body at λ→0 , Where λ - the length of the largest of the elementary segments ∆x k. Denote by V the volume of the given body, then by definition

On the other side,

Therefore, the volume of the body for given cross sections is calculated by the formula

If the body is formed by rotation around an axis Ox curvilinear trapezoid bounded from above by an arc of a continuous line y=f(x), Where a≤x≤b, That S(x)=πf 2 (x) and the last formula becomes:

Comment. The volume of a body obtained by rotating a curvilinear trapezoid bounded on the right by a function graph x=φ(y) (c ≤ x ≤ d), around the axis Oy calculated by the formula

Surface area of rotation

Consider the surface obtained by rotating the arc of the line y=f(x) (a≤x≤b) around the axis Ox(assume that the function y=f(x) has a continuous derivative). We fix the value x∈, the function argument will be incremented dx, which corresponds to the "elementary ring" obtained by rotating the elementary arc Δl. This "ring" is replaced by a cylindrical ring - the lateral surface of the body formed by the rotation of a rectangle with a base equal to the differential of the arc dl, and height h=f(x). Cutting the last ring and unfolding it, we get a strip with a width dl and length 2πy, Where y=f(x).

Therefore, the surface area differential is expressed by the formula

This formula expresses the surface area obtained by rotating the arc of a line y=f(x) (a≤x≤b) around the axis Ox.

Home > Lecture

Lecture 18. Applications of a definite integral.

18.1. Calculation of the areas of plane figures.

It is known that the definite integral on a segment is the area of a curvilinear trapezoid bounded by the graph of the function f(x). If the graph is located below the x-axis, i.e. f(x)< 0, то площадь имеет знак “-“, если график расположен выше оси Ох, т.е. f(x) >0, then the area has a “+” sign.

The formula is used to find the total area.

The area of a figure bounded by some lines can be found using certain integrals if the equations of these lines are known.

Example. Find the area of the figure bounded by the lines y \u003d x, y \u003d x 2, x \u003d 2.

The desired area (shaded in the figure) can be found by the formula:

18.2. Finding the area of a curvilinear sector.

To find the area of a curvilinear sector, we introduce a polar coordinate system. The equation of the curve that bounds the sector in this coordinate system has the form  = f(), where  is the length of the radius vector connecting the pole to an arbitrary point on the curve, and  is the angle of inclination of this radius vector to the polar axis.

The area of a curved sector can be found by the formula

18.3. Calculation of the arc length of a curve.

y y = f(x)

S i y i

The length of the polyline that corresponds to the arc can be found as
.

Then the length of the arc is
.

For geometric reasons:

In the same time

Then it can be shown that

Those.

If the equation of the curve is given parametrically, then, taking into account the rules for calculating the derivative of the parametrically given one, we obtain

where x = (t) and y = (t).

If set spatial curve, and x = (t), y = (t) and z = Z(t), then

If the curve is set to polar coordinates, That

,  = f().

Example: Find the circumference given by the equation x 2 + y 2 = r 2 .

1 way. Let us express the variable y from the equation.

Let's find the derivative

Then S = 2r. We got the well-known formula for the circumference of a circle.

2 way. If we represent the given equation in a polar coordinate system, we get: r 2 cos 2  + r 2 sin 2  = r 2, i.e. function  = f() = r,
Then

18.4. Calculation of volumes of bodies.

Calculation of the volume of a body from the known areas of its parallel sections.

Let there be a body of volume V. The area of any cross section of the body, Q, is known as a continuous function Q = Q(x). Let's divide the body into “layers” by cross sections passing through the points x i of the division of the segment . Because the function Q(x) is continuous on some intermediate segment of the partition, then it takes on its maximum and minimum values. Let's designate them accordingly M i and m i .

If on these largest and smallest sections to build cylinders with generators parallel to the x axis, then the volumes of these cylinders will be respectively equal to M i x i and m i x i here x i = x i - x i -1 .

Having made such constructions for all segments of the partition, we obtain cylinders whose volumes are, respectively,
And
.

As the partition step  tends to zero, these sums have a common limit:

Thus, the volume of the body can be found by the formula:

The disadvantage of this formula is that in order to find the volume, it is necessary to know the function Q(x), which is very problematic for complex bodies.

Example: Find the volume of a sphere of radius R.

In the cross sections of the ball, circles of variable radius y are obtained. Depending on the current x coordinate, this radius is expressed by the formula
.

Then the cross-sectional area function has the form: Q(x) =
.

We get the volume of the ball:

Example: Find the volume of an arbitrary pyramid with height H and base area S.

When crossing the pyramid with planes perpendicular to the height, in section we get figures similar to the base. The similarity coefficient of these figures is equal to the ratio x / H, where x is the distance from the section plane to the top of the pyramid.

It is known from geometry that the ratio of the areas of similar figures is equal to the coefficient of similarity squared, i.e.

From here we get the function of the cross-sectional areas:

Finding the volume of the pyramid:

18.5. The volume of bodies of revolution.

Consider the curve given by the equation y = f(x). Let us assume that the function f(x) is continuous on the segment . If the curvilinear trapezoid corresponding to it with bases a and b is rotated around the Ox axis, then we get the so-called body of revolution.

y = f(x)

Because each section of the body by the plane x = const is a circle of radius
, then the volume of the body of revolution can be easily found using the formula obtained above:

18.6. Surface area of a body of revolution.

M i B

Definition: Surface area of rotation curve AB around a given axis is the limit to which the areas of the surfaces of revolution of broken lines inscribed in the curve AB tend to, when the largest of the lengths of the links of these broken lines tend to zero.

Let's divide the arc AB into n parts by points M 0 , M 1 , M 2 , … , M n . The vertices of the resulting polyline have coordinates x i and y i . When rotating the broken line around the axis, we obtain a surface consisting of lateral surfaces of truncated cones, the area of which is equal to P i . This area can be found using the formula: