Definition: Let at each point of a smooth curve L=AB in the plane Oxy a continuous function of two variables is given f(x,y). Let's arbitrarily split the curve L on n parts with dots A = M 0, M 1, M 2, ... M n = B. Then on each of the resulting parts \(\bar((M)_(i-1)(M)_(i))\) we select any point \(\bar((M)_(i))\left(\ bar((x)_(i)),\bar((y)_(i))\right)\)and make the sum $$(S)_(n)=\sum_(i=1)^(n )f\left(\bar((x)_(i)),\bar((y)_(i))\right)\Delta (l)_(i)$$ where \(\Delta(l) _(i)=(M)_(i-1)(M)_(i)\) - arc of arc \(\bar((M)_(i-1)(M)_(i))\) . The amount received is called integral sum of the first kind for the function f(x,y) , given on the curve L.

Let us denote by d the largest of the arc lengths \(\bar((M)_(i-1)(M)_(i))\) (thus d = \(max_(i)\Delta(l)_(i)\ )). If at d? 0 there is a limit of integral sums S n (independent of the method of partitioning the curve L into parts and the choice of points \(\bar((M)_(i))\)), then this limit is called first order curvilinear integral from function f(x,y) along the curve L and is denoted by $$\int_(L)f(x,y)dl$$

It can be proven that if the function f(x,y) is continuous, then the line integral \(\int_(L)f(x,y)dl\) exists.

Properties of a curvilinear integral of the 1st kind

A curvilinear integral of the first kind has properties similar to the corresponding properties of a definite integral:

• additivity,
• linearity,
• module assessment,
• mean value theorem.

However, there is a difference: $$\int_(AB)f(x,y)dl=\int_(BA)f(x,y)dl$$ i.e. a line integral of the first kind does not depend on the direction of integration.

Calculation of curvilinear integrals of the first kind

The calculation of a curvilinear integral of the first kind is reduced to the calculation of a definite integral. Namely:

1. If the curve L is given by a continuously differentiable function y=y(x), x \(\in \) , then $$(\int\limits_L (f\left((x,y) \right)dl) ) = (\int \limits_a^b (f\left((x,y\left(x \right)) \right)\sqrt (1 + ((\left((y"\left(x \right)) \right))^ 2)) dx) ;)$$ in this case the expression \(dl=\sqrt((1 + ((\left((y"\left(x \right)) \right))^2))) dx \) is called the arc length differential.
2. If the curve L is specified parametrically, i.e. in the form x=x(t), y=y(t), where x(t), y(t) are continuously differentiable functions on some interval \(\left [ \alpha ,\beta \right ]\), then $$ (\int\limits_L (f\left((x,y) \right)dl) ) = (\int\limits_\alpha ^\beta (f\left ((x\left(t \right),y \left(t \right)) \right)\sqrt (((\left((x"\left(t \right)) \right))^2) + ((\left((y"\left(t \right)) \right))^2)) dt)) $$ This equality extends to the case of a spatial curve L defined parametrically: x=x(t), y=y(t), z=z(t), \(t\in \left [ \alpha ,\beta \right ]\). In this case, if f(x,y,z) is a continuous function along the curve L, then $$ (\int\limits_L (f\left((x,y,z) \right)dl) ) = (\int \limits_\alpha ^\beta (f\left [ (x\left(t \right),y\left(t \right),z\left(t \right)) \right ]\sqrt (((\left ((x"\left(t \right)) \right))^2) + ((\left((y"\left(t \right)) \right))^2) + ((\left(( z"\left(t \right)) \right))^2)) dt)) $$
3. If a plane curve L is given by the polar equation r=r(\(\varphi \)), \(\varphi \in\left [ \alpha ,\beta \right ] \), then $$ (\int\limits_L (f\ left((x,y) \right)dl) ) = (\int\limits_\alpha ^\beta (f\left((r\cos \varphi ,r\sin \varphi ) \right)\sqrt ((r ^2) + (((r)")^2)) d\varphi)) $$

Curvilinear integrals of the 1st kind - examples

Example 1

Calculate a line integral of the first kind

$$ \int_(L)\frac(x)(y)dl $$ where L is the arc of the parabola y 2 =2x, enclosed between points (2,2) and (8,4).

Solution: Find the differential of the arc dl for the curve \(y=\sqrt(2x)\). We have:

\((y)"=\frac(1)(\sqrt(2x)) \) $$ dl=\sqrt(1+\left ((y)" \right)^(2)) dx= \sqrt( 1+\left (\frac(1)(\sqrt(2x)) \right)^(2)) dx = \sqrt(1+ \frac(1)(2x)) dx $$ Therefore this integral is equal to: $ $\int_(L)\frac(x)(y)dl=\int_(2)^(8)\frac(x)(\sqrt(2x))\sqrt(1+\frac(1)(2x) )dx= \int_(2)^(8)\frac(x\sqrt(1+2x))(2x)dx= $$ $$ \frac(1)(2)\int_(2)^(8) \sqrt(1+2x)dx = \frac(1)(2).\frac(1)(3)\left (1+2x \right)^(\frac(3)(2))|_(2 )^(8)= \frac(1)(6)(17\sqrt(17)-5\sqrt(5)) $$

Example 2

Calculate the curvilinear integral of the first kind \(\int_(L)\sqrt(x^2+y^2)dl \), where L is the circle x 2 +y 2 =ax (a>0).

Solution: Let's introduce polar coordinates: \(x = r\cos \varphi \), \(y=r\sin \varphi \). Then since x 2 +y 2 =r 2, the equation of the circle has the form: \(r^(2)=arcos\varphi \), that is, \(r=acos\varphi \), and the differential of the arc $$ dl = \ sqrt(r^2+(2)"^2)d\varphi = $$ $$ =\sqrt(a^2cos^2\varphi=a^2sin^2\varphi )d\varphi=ad\varphi $$ .

In this case, \(\varphi\in \left [- \frac(\pi )(2) ,\frac(\pi )(2) \right ] \). Therefore, $$ \int_(L)\sqrt(x^2+y^2)dl=a\int_(-\frac(\pi )(2))^(\frac(\pi )(2))acos \varphi d\varphi =2a^2 $$

Purpose. The online calculator is designed to find the work done by force F when moving along the arc of a line L.

Curvilinear and surface integrals of the second kind

Consider the variety σ. Let τ(x,y,z) be the unit tangent vector to σ if σ is a curve, and let n(x,y,z) be the unit normal vector to σ if σ is a surface in R 3 . Let us introduce the vectors dl = τ · dl and dS = n · dS, where dl and dS are the length and area of ​​the corresponding section of the curve or surface. We will assume that dσ =dl if σ is a curve, and dσ =dS if σ is a surface. Let us call dσ the oriented measure of the corresponding section of the curve or surface.

Definition . Let an oriented continuous piecewise smooth manifold σ be given and a vector function on σ F(x,y,z)=P(x,y,z)i+Q(x,y,z)+R(x,y, z). Let's divide the manifold into parts with manifolds of lower dimension (a curve - with points, a surface - with curves), inside each resulting elementary manifold we choose a point M 0 (x 0 ,y 0 ,z 0), M 1 (x 1 ,y 1 ,z 1) , ... ,M n (x n ,y n ,z n). Let's count the values ​​of F(x i ,y i ,z i), i=1,2,...,n of the vector function at these points, scalarly multiply these values ​​by the oriented measure dσ i of the given elementary manifold (the oriented length or area of ​​the corresponding section of the manifold) and let's sum it up. The limit of the resulting sums, if it exists, does not depend on the method of dividing the manifold into parts and the choice of points inside each elementary manifold, provided that the diameter of the elementary section tends to zero, is called an integral over the manifold (a curvilinear integral if σ is a curve and a surface integral if σ - surface) of the second kind, an integral along an oriented manifold, or an integral of the vector F along σ, and is denoted in the general case, in the cases of curvilinear and surface integrals respectively.
Note that if F(x,y,z) is a force, then is the work of this force to move a material point along a curve, if F(x,y,z) is a stationary (time-independent) velocity field of a flowing fluid, then - the amount of liquid flowing through the surface S per unit time (vector flow through the surface).
If the curve is specified parametrically or, what is the same, in vector form,

That

and for the curvilinear integral of the second kind we have

Since dS = n dS =(cosα, cosβ, cosγ), where cosα, cosβ, cosγ are the direction cosines of the unit normal vector n and cosαdS=dydz, cosβdS=dxdz, cosγdS=dxdy, then for the surface integral of the second kind we obtain

If the surface is specified parametrically or, which is the same, in vector form
r(u,v)=x(u,v)i+y(u,v)j+z(u,v)k, (u,v)∈D
That

Where - Jacobians (determinants of Jacobi matrices, or, what is the same, matrices of derivatives) of vector functions respectively.

If the surface S can be specified simultaneously by equations, then the surface integral of the second kind is calculated by the formula

where D 1, D 2, D 3 are the projections of the surface S onto the coordinate planes Y0Z, X0Z, X0Y, respectively, and the “+” sign is taken if the angle between the normal vector and the axis along which the design is carried out is acute, and the “–” sign, if this angle is obtuse.

Properties of curvilinear and surface integrals of the second kind

Let us note some properties of curvilinear and surface integrals of the second kind.
Theorem 1. Curvilinear and surface integrals of the 2nd kind depend on the orientation of the curve and surface, more precisely
.

Theorem 2. Let σ=σ 1 ∪σ 2 and the dimension of the intersection dlim(σ 1 ∩σ 2)=n-1. Then

Proof. By including the common boundary σ 1 with σ 2 among the partition manifolds in the definition of the integral over a manifold of the second kind, we obtain the required result.

Example No. 1. Find the work done by force F when moving along the arc of line L from point M 0 to point M 1.
F=x 2 yi+yj; , L: segment M 0 M 1
M 0 (-1;3), M 0 (0;1)
Solution.
Find the equation of a straight line along the segment M 0 M 1 .
or y=-2x+1
dy=-2dx

Limits of change x: [-1; 0]

A curvilinear integral of the 2nd kind is calculated in the same way as a curvilinear integral of the 1st kind by reduction to the definite. To do this, all variables under the integral sign are expressed through one variable, using the equation of the line along which the integration is performed.

a) If the line AB is given by a system of equations then

(10.3)

For the plane case, when the curve is given by the equation the curvilinear integral is calculated using the formula: . (10.4)

If the line AB is given by parametric equations then

(10.5)

For a flat case, if the line AB given by parametric equations , the curvilinear integral is calculated by the formula:

, (10.6)

where are the parameter values t, corresponding to the starting and ending points of the integration path.

If the line AB piecewise smooth, then we should use the property of additivity of the curvilinear integral by splitting AB on smooth arcs.

Example 10.1 Let's calculate the curvilinear integral along a contour consisting of part of a curve from point before and ellipse arcs from point before .

Since the contour consists of two parts, we use the additivity property of the curvilinear integral: . Let us reduce both integrals to definite ones. Part of the contour is given by an equation relative to the variable . Let's use the formula (10.4 ), in which we switch the roles of the variables. Those.

. After calculation we get .

To calculate the contour integral Sun Let's move on to the parametric form of writing the ellipse equation and use formula (10.6).

Pay attention to the limits of integration. Point corresponds to the value, and to the point corresponds Answer:
.

Example 10.2. Let's calculate along a straight line segment AB, Where A(1,2,3), B(2,5,8).

Solution. A curvilinear integral of the 2nd kind is given. To calculate it, you need to convert it to a specific one. Let's compose the equations of the line. Its direction vector has coordinates .

Canonical equations of line AB: .

Parametric equations of this line: ,

At
.

Let's use the formula (10.5) :

Having calculated the integral, we get the answer: .

5. Work of force when moving a material point of unit mass from point to point along a curve .

Let at each point of a piecewise smooth curve a vector is given that has continuous coordinate functions: . Let's break this curve into small parts with points so that at the points of each part meaning of functions
could be considered constant, and the part itself could be mistaken for a straight segment (see Fig. 10.1). Then . The scalar product of a constant force, the role of which is played by a vector , per rectilinear displacement vector is numerically equal to the work done by the force when moving a material point along . Let's make an integral sum . In the limit, with an unlimited increase in the number of partitions, we obtain a curvilinear integral of the 2nd kind

. (10.7) Thus, the physical meaning of the curvilinear integral of the 2nd kind - this is work done by force when moving a material point from A To IN along the contour L.

Example 10.3. Let's calculate the work done by the vector when moving a point along a portion of a Viviani curve defined as the intersection of a hemisphere and cylinder , running counterclockwise when viewed from the positive part of the axis OX.

Solution. Let's construct the given curve as the line of intersection of two surfaces (see Fig. 10.3).

.

To reduce the integrand to one variable, let’s move to a cylindrical coordinate system: .

Because a point moves along a curve , then it is convenient to choose as a parameter a variable that changes along the contour so that . Then we obtain the following parametric equations of this curve:

.Wherein
.

Let us substitute the resulting expressions into the formula for calculating circulation:

( - the + sign indicates that the point moves along the contour counterclockwise)

Let's calculate the integral and get the answer: .

Lesson 11.

Green's formula for a simply connected region. Independence of the curvilinear integral from the path of integration. Newton-Leibniz formula. Finding a function from its total differential using a curvilinear integral (plane and spatial cases).

OL-1 chapter 5, OL-2 chapter 3, OL-4 chapter 3 § 10, clause 10.3, 10.4.

Practice : OL-6 No. 2318 (a, b, d), 2319 (a, c), 2322 (a, d), 2327, 2329 or OL-5 No. 10.79, 82, 133, 135, 139.

Home building for lesson 11: OL-6 No. 2318 (c, d), 2319 (c, d), 2322 (b, c), 2328, 2330 or OL-5 No. 10.80, 134, 136, 140

Green's formula.

Let on the plane given a simply connected domain bounded by a piecewise smooth closed contour. (A region is called simply connected if any closed contour in it can be contracted to a point in this region).

Theorem. If the functions and their partial derivatives G, That

Figure 11.1

- Green's formula . (11.1)

Indicates positive bypass direction (counterclockwise).

Example 11.1. Using Green's formula, we calculate the integral along a contour consisting of segments O.A., O.B. and greater arc of a circle , connecting the points A And B, If , , .

Solution. Let's build a contour (see Fig. 11.2). Let us calculate the necessary derivatives.

Figure 11.2

, ; , . Functions and their derivatives are continuous in a closed region bounded by a given contour. According to Green's formula, this integral is .

After substituting the calculated derivatives we get

. We calculate the double integral by moving to polar coordinates:
.

Let's check the answer by calculating the integral directly along the contour as a curvilinear integral of the 2nd kind.
.

Answer:
.

2. Independence of the curvilinear integral from the path of integration.

Let And - arbitrary points of a simply connected region pl. . Curvilinear integrals calculated from different curves connecting these points generally have different meanings. But if certain conditions are met, all these values ​​may turn out to be the same. Then the integral does not depend on the shape of the path, but depends only on the starting and ending points.

The following theorems hold.

Theorem 1. In order for the integral
did not depend on the shape of the path connecting the points and , it is necessary and sufficient that this integral along any closed contour be equal to zero.

Theorem 2.. In order for the integral
along any closed contour is equal to zero, it is necessary and sufficient that the function and their partial derivatives were continuous in a closed region G and so that the condition is satisfied (11.2)

Thus, if the conditions for the integral to be independent of the path shape are met (11.2) , then it is enough to specify only the start and end points: (11.3)

Theorem 3. If the condition is satisfied in a simply connected region , then there is a function such that . (11.4)

This formula is called formula Newton–Leibniz for the line integral.

Comment. Recall that the equality is a necessary and sufficient condition for the fact that the expression
.

Then from the above theorems it follows that if the functions and their partial derivatives continuous in a closed region G, in which the points are given And , And , That

a) there is a function , such that ,

does not depend on the shape of the path, ,

c) the formula holds Newton–Leibniz .

Example 11.2. Let us make sure that the integral
does not depend on the shape of the path, and let's calculate it.

Solution. .

Figure 11.3

Let's check that condition (11.2) is satisfied.
. As we can see, the condition is met. The value of the integral does not depend on the path of integration. Let us choose the integration path. Most

a simple way to calculate is a broken line DIA, connecting the starting and ending points of a path. (See Fig. 11.3)

Then .

3. Finding a function by its total differential.

Using a curvilinear integral, which does not depend on the shape of the path, we can find the function , knowing its full differential. This problem is solved as follows.

If the functions and their partial derivatives continuous in a closed region G And , then the expression is the total differential of some function . In addition, the integral
, firstly, does not depend on the shape of the path and, secondly, can be calculated using the Newton–Leibniz formula.

Let's calculate
two ways.

Figure 11.4

a) Select a point in the region with specific coordinates and point with arbitrary coordinates. Let us calculate the curvilinear integral along a broken line consisting of two line segments connecting these points, with one of the segments parallel to the axis and the other to the axis. Then . (See Fig. 11.4)

The equation .

The equation .

We get: Having calculated both integrals, we get a certain function in the answer .

b) Now we calculate the same integral using the Newton–Leibniz formula.

Now let's compare two results of calculating the same integral. Functional part the answer in point a) is the desired function , and the numerical part is its value at the point .

Example 11.3. Let's make sure that the expression
is the total differential of some function and we'll find her. Let's check the results of calculating example 11.2 using the Newton-Leibniz formula.

Solution. Condition for the existence of a function (11.2) was checked in the previous example. Let's find this function, for which we will use Figure 11.4, and take for point . Let's compose and calculate the integral along the broken line DIA, Where :

As mentioned above, the functional part of the resulting expression is the desired function
.

Let's check the result of the calculations from Example 11.2 using the Newton–Leibniz formula:

The results were the same.

Comment. All the statements considered are also true for the spatial case, but with a larger number of conditions.

Let a piecewise smooth curve belong to a region in space . Then, if the functions and their partial derivatives are continuous in the closed region in which the points are given And , And
(11.5 ), That

a) the expression is the total differential of some function ,

b) curvilinear integral of the total differential of some function does not depend on the shape of the path and ,

c) the formula holds Newton–Leibniz .(11.6 )

Example 11.4. Let's make sure that the expression is the complete differential of some function and we'll find her.

Solution. To answer the question of whether a given expression is a complete differential of some function , let's calculate the partial derivatives of the functions, ,
. (Cm. (11.5) ) ; ; ; ; ; .

These functions are continuous along with their partial derivatives at any point in space .

We see that the necessary and sufficient conditions for existence are satisfied : , , , etc.

To calculate a function Let us take advantage of the fact that the linear integral does not depend on the path of integration and can be calculated using the Newton-Leibniz formula. Let the point - the beginning of the path, and some point - end of the road . Let's calculate the integral

along a contour consisting of straight segments parallel to the coordinate axes. (see Fig. 11.5).

.

Figure 11.5

Equations of the contour parts: , ,
.

Then

, x fixed here, so ,

, recorded here y, That's why .

As a result we get: .

Now let's calculate the same integral using the Newton-Leibniz formula.

Let's compare the results: .

From the resulting equality it follows that , and

Lesson 12.

Surface integral of the first kind: definition, basic properties. Rules for calculating a surface integral of the first kind using a double integral. Applications of the surface integral of the first kind: surface area, mass of a material surface, static moments about coordinate planes, moments of inertia, and coordinates of the center of gravity. OL-1 ch.6, OL 2 ch.3, OL-4§ 11.

Practice: OL-6 No. 2347, 2352, 2353 or OL-5 No. 10.62, 65, 67.

Homework for lesson 12:

OL-6 No. 2348, 2354 or OL-5 No. 10.63, 64, 68.

It is more convenient to calculate volume in cylindrical coordinates. Equation of a circle bounding a region D, a cone and a paraboloid

respectively take the form ρ = 2, z = ρ, z = 6 − ρ 2. Taking into account the fact that this body is symmetrical relative to the xOz and yOz planes. we have

	6− ρ 2
V = 4 ∫ 2 dϕ ∫ ρ dρ ∫ dz = 4 ∫ 2 dϕ ∫ ρ z				6 ρ − ρ 2 d ρ =

4 ∫ d ϕ∫ (6 ρ − ρ3 − ρ2 ) d ρ =

2 d ϕ =

4 ∫ 2 (3 ρ 2 −

∫ 2 d ϕ =

32π

If symmetry is not taken into account, then

6− ρ 2

32π

V = ∫

dϕ ∫ ρ dρ ∫ dz =

3. CURVILINEAR INTEGRALS

Let us generalize the concept of a definite integral to the case when the domain of integration is a certain curve. Integrals of this kind are called curvilinear. There are two types of curvilinear integrals: curvilinear integrals along the length of the arc and curvilinear integrals over the coordinates.

3.1. Definition of a curvilinear integral of the first type (along the length of the arc). Let the function f(x,y) defined along a flat piecewise

smooth1 curve L, the ends of which will be points A and B. Let us divide the curve L arbitrarily into n parts with points M 0 = A, M 1,... M n = B. On

For each of the partial arcs M i M i + 1, we select an arbitrary point (x i, y i) and calculate the values ​​of the function f (x, y) at each of these points. Sum

1 A curve is called smooth if at each point there is a tangent that continuously changes along the curve. A piecewise smooth curve is a curve consisting of a finite number of smooth pieces.

n− 1
σ n = ∑ f (x i , y i ) ∆ l i ,

i = 0

where ∆ l i is the length of the partial arc M i M i + 1, called integral sum

for the function f(x, y) along the curve L. Let us denote the largest of the lengths
partial arcs M i M i + 1 , i =
	0 ,n − 1 through λ , that is, λ = max ∆ l i .
		0 ≤i ≤n −1
If there is a finite limit I of the integral sum (3.1)
tending to zero of the largest of the lengths of partial arcsM i M i + 1,
depending neither on the method of dividing the curve L into partial arcs, nor on

choice of points (x i, y i), then this limit is called curvilinear integral of the first type (curvilinear integral along the length of the arc) from the function f (x, y) along the curve L and is denoted by the symbol ∫ f (x, y) dl.

Thus, by definition
n− 1
I = lim ∑ f (xi , yi ) ∆ li = ∫ f (x, y) dl.
λ → 0 i = 0

The function f(x, y) is called in this case integrable along the curve L,

the curve L = AB is the contour of integration, A is the initial point, and B is the final point of integration, dl is the element of arc length.

Remark 3.1. If in (3.2) we put f (x, y) ≡ 1 for (x, y) L, then

we obtain an expression for the length of the arc L in the form of a curvilinear integral of the first type

l = ∫ dl.

Indeed, from the definition of a curvilinear integral it follows that
	dl = lim n − 1
		∆l	Lim l = l .
	λ → 0 ∑		λ→ 0
	i = 0
3.2. Basic properties of the first type of curvilinear integral
are similar to the properties of a definite integral:
1 o. ∫ [ f1 (x, y) ± f2 (x, y) ] dl = ∫ f1 (x, y) dl ± ∫ f2 (x, y) dl.
2 o. ∫ cf (x, y) dl = c ∫ f (x, y) dl, where c is a constant.
				and L, not
3 o. If the integration loop L is divided into two parts L

having common interior points, then

∫ f (x, y)dl = ∫ f (x, y)dl + ∫ f (x, y)dl.

4 o. We especially note that the value of the curvilinear integral of the first type does not depend on the direction of integration, since the values ​​of the function f (x, y) in

arbitrary points and the length of partial arcs ∆ l i , which are positive,

regardless of which point of the curve AB is considered the initial and which is the final, that is

	f (x, y) dl = ∫ f (x, y) dl .
3.3. Calculation of a curve integral of the first type
reduces to calculating definite integrals.
	x= x(t)
Let the curve L given by parametric equations		y=y(t)

Let α and β be the values ​​of the parameter t corresponding to the beginning (point A) and

end (point B)

[α , β ]

x(t), y(t) and

derivatives

x (t), y (t)

Continuous

f(x, y) -

is continuous along the curve L. From the course of differential calculus

functions of one variable it is known that

dl = (x(t))

+ (y(t))

∫ f (x, y) dl = ∫ f (x(t), y(t))

(x(t)

+ (y(t))

∫ x2 dl,

Example 3.1.

Calculate

circle

x= a cos t

0 ≤ t ≤

y= a sin t

Solution. Since x (t) = − a sin t, y (t) = a cos t, then

dl =

(− a sin t) 2 + (a cos t) 2 dt = a2 sin 2 t + cos 2 tdt = adt

and from formula (3.4) we obtain

Cos 2t )dt =

sin 2t

∫ x2 dl = ∫ a2 cos 2 t adt = a

3 ∫

πa 3

sinπ

L is given

equation

y = y(x) ,

a ≤ x ≤ b

y(x)

is continuous along with its derivative y

(x) for a ≤ x ≤ b, then

dl =

1+(y(x))

and formula (3.4) takes the form

∫ f (x, y) dl = ∫ f (x, y(x))

(y(x))

L is given

x = x(y), c ≤ y ≤ d

x(y)

equation

is continuous along with its derivative x (y) for c ≤ y ≤ d, then

dl =

1+(x(y))

and formula (3.4) takes the form

∫ f (x, y) dl = ∫ f (x(y), y)

1 + (x(y))

Example 3.2. Calculate ∫ ydl, where L is the arc of the parabola

2 x from

point A (0,0) to point B (2,2).

Solution . Let's calculate the integral in two ways, using

formulas (3.5) and (3.6)

1) Let's use formula (3.5). Because

2x (y ≥ 0), y ′

2 x =

2 x

dl =

1+ 2 x dx,

3 / 2 2

1 (5

3 2 − 1) .

∫ ydl = ∫

2 x + 1 dx = ∫ (2 x + 1) 1/ 2 dx =

1 (2x + 1)

2) Let's use formula (3.6). Because

x = 2 , x

Y, dl

1 + y

		y 1 + y 2 dy =		(1 + y				/ 2 2
∫ ydl = ∫
				3 / 2

1 3 (5 5 − 1).

Remark 3.2. Similar to what was considered, we can introduce the concept of a curvilinear integral of the first type of the function f (x, y, z) over

spatial piecewise smooth curve L:

If the curve L is given by parametric equations

α ≤ t ≤ β, then

dl =

(x(t))

(y(t))

(z(t))

∫ f (x, y, z) dl =

= ∫

dt.

f (x (t), y (t), z (t)) (x (t))

(y(t))

(z(t))

x= x(t) , y= y(t)

z= z(t)

Example 3.3. Calculate∫ (2 z − x 2 + y 2 ) dl , where L is the arc of the curve

x= t cos t	0 ≤ t ≤ 2 π.
y = t sin t
z = t
	x′ = cost − t sint, y′ = sint + t cost, z′ = 1 ,
	dl =	(cos t − t sin t)2 + (sin t + t cos t)2 + 1 dt =

Cos2 t − 2 t sin t cos t + t2 sin2 t + sin2 t + 2 t sin t cos t + t2 cos2 t + 1 dt =

2 + t2 dt .

Now, according to formula (3.7), we have

∫ (2z −

x2 + y2 ) dl = ∫ (2 t −

t 2 cos 2 t + t 2 sin 2 t )

2 + t 2 dt =

T2)

= ∫

t2+t

dt =

4π

− 2 2

cylindrical

surfaces,

which is made up of perpendiculars to

xOy plane,

restored at points

(x, y)

L=AB

and having

represents the mass of a curve L having a variable linear density ρ(x, y)

the linear density of which varies according to the law ρ (x, y) = 2 y.

Solution. To calculate the mass of the arc AB, we use formula (3.8). The arc AB is given parametrically, so to calculate the integral (3.8) we use formula (3.4). Because

1+t

dt,

x (t) = 1, y (t) = t, dl =

3/ 2 1

1 (1+ t

m = ∫ 2 ydl = ∫

1 2 + t2 dt = ∫ t 1 + t2 dt =

(2 3 / 2 −

1) =

2 2 − 1.

3.4. Definition of a curvilinear integral of the second type (by

coordinates). Let the function

f(x, y) is defined along a plane

piecewise smooth curve L, the ends of which will be points A and B. Again

arbitrary

let's break it

curve L

M 0 = A , M 1 ,... M n = B We also choose within

each partial

arcs M i M i + 1

arbitrary point

(xi, yi)

and calculate

A curve AB defined by parametric equations is called smooth if the functions and have continuous derivatives on the segment, and if at a finite number of points on the segment these derivatives do not exist or simultaneously vanish, then the curve is called piecewise smooth. Let AB be a flat curve, smooth or piecewise smooth. Let f(M) be a function defined on the curve AB or in some domain D containing this curve. Let's consider the division of the curve A B into parts by points (Fig. 1). Let us choose an arbitrary point Mk on each of the arcs A^At+i and compose a sum where Alt is the length of the arc and call it the integral sum for the function f(M) over the length of the arc of the curve. Let D / be the largest of the lengths of partial arcs, i.e. Properties of curvilinear integrals of the 1st kind for space curves Curvilinear integrals of the 2nd kind Calculation of a curvilinear integral Properties Relationship between Definitions. If at the integral sum (I) has a finite limit that does not depend either on the method of partitioning the curve AB into parts or on the choice of points on each of the arcs of the partition, then this limit is called a curvilinear integral of the \th kind of the function f(M) over the curve AB (the integral over the length of the arc of the curve) and is denoted by the symbol In this case, the function /(M) is called integrable along the curve ABU, the curve A B is called the contour of integration, A is the initial point, B is the end point of integration. Thus, by definition, Example 1. Let a mass with variable linear density J(M) be distributed along some smooth curve L. Find the mass m of curve L. (2) Let us divide the curve L into n arbitrary parts) and calculate approximately the mass of each part, assuming that on each part the density is constant and equal to the density at any of its points, for example, at the extreme left point /(Af*). Then the sum ksh where D/d is the length of the Dth part, will be an approximate value of the mass m. It is clear that the smaller the partition of the curve L, the smaller the error. we obtain the exact value of the mass of the entire curve L, i.e. But the limit on the right is a curvilinear integral of the 1st kind. So, 1.1. Existence of a curvilinear integral of the 1st kind Let us take as a parameter on the curve AB the length of the arc I, measured from the starting point A (Fig. 2). Then the AB curve can be described by equations (3) where L is the length of the AB curve. Equations (3) are called natural equations of the AB curve. When passing to natural equations, the function f(x) y), defined on the curve AB, will be reduced to a function of the variable I: / (x(1)) y(1)). Having denoted by the value of the parameter I corresponding to the point Mky, we rewrite the integral sum (I) in the form This is the integral sum corresponding to a certain integral. Since the integral sums (1) and (4) are equal to each other, then the integrals corresponding to them are equal. Thus, (5) Theorem 1. If the function /(M) is continuous along a smooth curve AB, then there is a curvilinear integral (since under these conditions there is a definite integral on the right in equality (5). 1.2. Properties of curvilinear integrals of the 1st kind 1. From the form of the integral sum (1) it follows that i.e. the value of a curvilinear integral of the 1st kind does not depend on the direction of integration. 2. Linearity. If for each of the functions /() there is a curvilinear integral along the curve ABt, then for the function a/, where a and /3 are any constants, there also exists a curvilinear integral along the curve AB> and 3. Additivity. If the curve AB consists of two pieces and for the function /(M) there is a curvilinear integral over ABU, then there are integrals with 4. If 0 on the curve AB, then 5. If the function is integrable on the curve AB, then the function || is also integrable on A B, and at the same time b. Average formula. If the function / is continuous along the curve AB, then on this curve there is a point Mc such that where L is the length of the curve AB. 1.3. Calculation of a curvilinear integral of the 1st kind Let the curve AB be given by parametric equations, with point A corresponding to the value t = to, and point B to the value. We will assume that the functions) are continuous on together with their derivatives and the inequality is satisfied. Then the differential of the arc of the curve is calculated by the formula. In particular, if the curve AB is given by an explicit equation is continuously differentiable on [a, b] and the point A corresponds to the value x = a, and the point B - value x = 6, then, taking x as a parameter, we get 1.4. Curvilinear integrals of the 1st kind for spatial curves The definition of a curvilinear integral of the 1st kind, formulated above for a plane curve, is literally carried over to the case when the function f(M) is given along some spatial curve AB. Let the curve AB be given by parametric equations Properties of curvilinear integrals of the 1st kind for spatial curves Curvilinear integrals of the 2nd kind Calculation of a curvilinear integral Properties Relationship between Then the curvilinear integral taken along this curve can be reduced to a definite integral using the following formula: Example 2. Calculate the curvilinear integral where L is the contour of a triangle with vertices at a point* (Fig. 3). By the property of additivity we have Let us calculate each of the integrals separately. Since on the segment OA we have: , then on the segment AN we have, where and then Fig. Finally, Therefore, Note. When calculating the integrals, we used property 1, according to which. Curvilinear integrals of the 2nd kind Let A B be a smooth or piecewise smooth oriented curve on the xOy plane and let be a vector function defined in some domain D containing the curve AB. Let us divide the curve AB into parts with points whose coordinates we denote respectively by (Fig. 4). On each of the elementary arcs AkAk+\ we take an arbitrary point and make a sum. Let D/ be the length of the largest of the arcs. Definition. If at sum (1) has a finite limit that does not depend on either the method of partitioning the curve AB or the choice of points rjk) on elementary arcs, then this limit is called the curvilinear integral of the 2-city of the vector function along the curve AB and is denoted by the symbol So by definition Theorem 2. If in some domain D containing the curve AB the functions are continuous, then the curvilinear integral of the 2-city exists. Let be the radius vector of the point M(x, y). Then the integrand in formula (2) can be represented as a scalar product of the vectors F(M) and dr. So the integral of the 2nd kind of a vector function along the curve AB can be written briefly as follows: 2.1. Calculation of a curvilinear integral of the 2nd kind Let the curve AB be defined by parametric equations, where the functions are continuous along with the derivatives on the segment, and a change in the parameter t from t0 to t\ corresponds to the movement of a point along the curve AB of point A to point B. If in some region D, containing the curve AB, the functions are continuous, then the curvilinear integral of the 2nd kind is reduced to the following definite integral: Thus, the calculation of the curvilinear integral of the 2nd kind can also be reduced to the calculation of the definite integral. O) Example 1. Calculate the integral along a straight line segment connecting the points 2) along a parabola connecting the same points) Equation of a line parameter, whence So 2) Equation of line AB: Hence therefore The considered example anoints that the value of a curved integral of the 2nd kind , generally speaking, depends on the shape of the integration path. 2.2. Properties of a curvilinear integral of the 2nd kind 1. Linearity. If there are Properties of curvilinear integrals of the 1st kind for space curves Curvilinear integrals of the 2nd kind Calculation of a curvilinear integral Properties The connection between then for any real a and /5 there is an integral where 2. Additenost. If the curve AB is divided into parts AC and SB and a curvilinear integral exists, then integrals also exist. The last property of the physical interpretation of a curvilinear integral of the 2nd kind is the work of the force field F along a certain path: when the direction of deshkeniya along the curve changes, the work of the force field along this curve changes sign to the opposite. 2.3. Relationship between curvilinear integrals of the 1st and 2nd kind Consider a curvilinear integral of the 2nd kind where the oriented curve AB (A is the starting point, B is the end point) is given by the vector equation (here I is the length of the curve, measured in the direction in which the AB curve is oriented) (Fig. 6). Then dr or where r = m(1) is the unit vector of the tangent to the curve AB at point M(1). Then Note that the last integral in this formula is a curvilinear integral of the 1st kind. When the orientation of the curve AB changes, the unit vector of the tangent r is replaced by the opposite vector (-r), which entails a change in the sign of its integrand and, therefore, the sign of the integral itself. Site navigation Health Psychology Relationship Job Entertainment Latest articles Thermal phenomena presentation Ancient peoples' ideas about the Universe Who is Claude Shannon and why is he famous? The accession of Darius Kodoman Brief description of Darius 3 commander The difference between a hydrogen bomb and an atomic bomb: a list of differences, history of creation