Integration path independence for total differential. Conditions for the independence of a curvilinear integral from the path of integration on the plane

Consider the curvilinear integral

taken along some plane curve L connecting points M and N. We will assume that the functions have continuous partial derivatives in the region D under consideration. Let us find out under what conditions the written curvilinear integral does not depend on the shape of the curve L, but depends only on the position of the initial and final points M and N.

Let us consider two arbitrary curves MPN and MQN, lying in the considered region D and connecting points M and N (Fig. 351). Let

Then, based on properties 1 and 2 of curvilinear integrals (§ 1), we have

i.e. line integral over a closed loop

In the last formula, the line integral is taken over a closed contour L composed of curves. This contour L can obviously be considered arbitrary.

Thus, from the condition that for any two points M and N the line integral does not depend on the shape of the curve connecting them, but depends only on the position of these points, it follows that the line integral along any closed contour is equal to zero.

The converse conclusion is also true: if a curvilinear integral over any closed contour is equal to zero, then this curvilinear integral does not depend on the shape of the curve connecting any two points, but depends only on the position of these points. Indeed, equality (2) implies equality (1).

In example 4 of § 2, the curvilinear integral does not depend on the path of integration; in example 3, the curvilinear integral depends on the path of integration, since in this example the integral over a closed contour does not equal zero, but gives the area limited by the contour in question; in examples 1 and 2, the line integrals also depend on the path of integration.

The question naturally arises: what conditions must the functions satisfy in order for the curvilinear integral along any closed contour to be equal to zero. The answer to this question is given by the following theorem:

Theorem. Let the functions, together with their partial derivatives, be continuous at all points of some domain D. Then, in order for the curvilinear integral over any closed contour L lying in the region D to be equal to zero, i.e., so that

it is necessary and sufficient for the equality to be fulfilled

in all heats of the region

Proof. Let us consider an arbitrary closed contour L in the domain D and write Green’s formula for it:

If condition (3) is satisfied, then the double integral on the left is identically equal to zero and, therefore,

Thus, the sufficiency of condition (3) is proven.

Let us now prove the necessity of this condition, i.e. we will prove that if equality (2) is satisfied for any closed curve L in the domain D, then at each point of this domain condition (3) is also satisfied.

Let us assume, on the contrary, that equality (2) is satisfied, i.e.

and condition (3) is not satisfied, i.e.

at least at one point. Let, for example, at some point we have the inequality

Since the left side of the inequality contains a continuous function, it will be positive and greater than a certain number at all points of some sufficiently small domain D containing the point . Let's take the double integral over this area of ​​the difference. It will have a positive meaning. Really,

But according to Green's formula, the left-hand side of the last inequality is equal to the curvilinear integral along the boundary of the region, which is equal to zero. Consequently, the last inequality contradicts condition (2) and, therefore, the assumption that it is different from zero at least at one point is incorrect. From here

it follows that

at all points in this area

Thus, the theorem is completely proven.

In § 9 ch. XIII it was proven that the fulfillment of the condition is equivalent to the fact that the expression is the total differential of some function, i.e.

But in this case the vector

there is a gradient of a function whose gradient is equal to a vector is called the potential of this vector. Let us prove that in this case the curvilinear integral

For any curve L connecting points M and N, (M) is equal to the difference between the values ​​of the function and at these points:

Proof. If is the complete differential of the function, then the curvilinear integral will take the form

To calculate this integral, we write the parametric equations of the curve L connecting points M and

integral will be reduced to the following definite integral:

The expression in parentheses is a function of which is a complete derivative of the function Therefore

As we see, the line integral of the total differential does not depend on the shape of the curve along which the integration is performed.

A similar statement holds for a curvilinear integral over a space curve (see § 7 below).

Comment. Sometimes it is necessary to consider curvilinear integrals along the arc length L of some function

A region is called simply connected if its boundary is a connected set. A region is called n-connected if its boundary splits into n-connected sets.

Comment. Green's formula is also true for multiply connected regions.

In order for the integral (A, B – any points from D) to not depend on the path of integration (but only on the initial and final points A, B), it is necessary and sufficient that along any closed curve (along any contour) lying in D the integral was equal to zero =0

Proof (necessity). Let (4) be independent of the integration path. Consider an arbitrary contour C lying in the region D and choose two arbitrary points A, B on this contour. Then curve C can be represented as the union of two curves AB=G2, AB=G1, C=Г - 1 + G2.

Theorem 1. In order for a curvilinear integral to be independent of the path of integration in D, it is necessary and sufficient that

in area D. Sufficiency. If this is true, then Green’s formula for any contour C will be whence the required statement follows by the lemma. Necessity. By lemma for any contour = 0. Then, by Green's formula for the area D bounded by this contour = 0. By the mean value theorem = mD or = = 0. Passing to the limit, contracting the contour to a point, we obtain that at this point.

Theorem 2. In order for the curvilinear integral (4) to be independent of the path of integration in D, it is necessary and sufficient that the integrand expression Pdx+Qdy be the total differential of some function u in the domain D. du = Pdx+Qdy. Adequacy. Let it be fulfilled, then Necessity. Let the integral be independent of the path of integration. We fix some point A0 in the domain D and define the function u(A) = u(x,y)=

In this case

XО (xО). Thus, there is a derivative =P. Similarly, it is checked that =Q. Under the assumptions made, the function u turns out to be continuously differentiable and du = Pdx+Qdy.

32-33. Definition of curvilinear integrals of the 1st and 2nd kind

Curvilinear integral over arc length (1st kind)

Let the function f(x,y) be defined and continuous at the points of the arc AB of a smooth curve K. Arbitrarily divide the arc into n elementary arcs by points t0..tn let lk be the length k of the particular arc. Let us take an arbitrary point N(k,k) on each elementary arc and multiply this point by the corresponding point. the length of the arc will be composed of three integral sums:

1 =f(k,k)lk 2 = Р(k,k)хk 3 = Q(k,k)yk, where хk = x k -x k -1 , yk = y k -y k -1

The curvilinear integral of the 1st kind along the length of the arc will be called the limit of the integral sum 1, provided that max(lk)  0

If the limit of the integral sum is 2 or 3 at   0, then this limit is called. curvilinear integral of the 2nd kind, a function P(x,y) or Q(x,y) along the curve l = AB and is denoted:
or

amount:
+
It is customary to call it a general curvilinear integral of the 2nd kind and denote it by the symbol:
in this case, the functions f(x,y), P(x,y), Q(x,y) are called integrable along the curve l = AB. The curve l itself is called a contour or by integration A is the initial point, B is the final integration point, dl is the differential of the arc length, therefore the curvilinear integral of the 1st kind is called. a curvilinear integral over an arc of a curve, and of the second kind – over a function..

From the definition of curvilinear integrals it follows that integrals of the 1st kind do not depend on the direction in which the curve l is run from A and B or from B and A. Curvilinear integral of the 1st kind along AB:

, for curvilinear integrals of the 2nd kind, a change in the direction of the curve leads to a change in sign:

In the case when l is a closed curve, i.e., point B coincides with point A, then of the two possible directions for traversing the closed contour, l is called positive the direction in which the area lying inside the contour remains to the left with respect to??? making a round, i.e. the direction of movement is counterclockwise. The opposite direction of traversal is called negative. The curvilinear integral AB along a closed contour l traversed in the positive direction will be denoted by the symbol:

For a spatial curve, one integral of the 1st kind is similarly introduced:

and three integrals of the 2nd kind:

the sum of the last three integrals is called. general curvilinear integral of the 2nd kind.

Some applications of curvilinear integrals of the 1st kind.

1.Integral
- arc length AB

2.Mechanical meaning of the integral of the 1st kind.

If f(x,y) = (x,y) is the linear density of the material arc, then its mass:

3.Coordinates of the center of mass of the material arc:

4. Moment of inertia of an arc lying in the oxy plane relative to the origin of coordinates and the axes of rotation ox, oy:

5. Geometric meaning of the integral of the 1st kind

Let the function z = f(x,y) – have the dimension of length f(x,y)>=0 at all points of the material arc lying in the oxy plane then:

, where S is the area of ​​the cylindrical surface, the cat consists of perpendiculars to the okha plane, east. at points M(x,y) of the AB curve.

Some applications of curvilinear integrals of the 2nd kind.

Calculation of the area of ​​a flat region D with boundary L

2. Work of force. Let a material point, under the influence of a force, move along a continuous flat curve BC, heading from B to C, the work of this force is:

From the path of integration.

Consider a curvilinear integral of the 2nd kind, where L– a curve connecting points M And N. Let the functions P(x, y) And Q(x, y) have continuous partial derivatives in some domain D, which contains the entire curve L. Let us determine the conditions under which the curvilinear integral under consideration does not depend on the shape of the curve L, but only on the location of the points M And N.

Let's draw two arbitrary curves MPN And MQN, lying in the area D and connecting points M And N(Fig. 1).

Q

M N Rice. 1.

Let's pretend that , that is

Then where L– a closed contour made up of curves MPN And N.Q.M.(hence, it can be considered arbitrary). Thus, the condition for the independence of a curvilinear integral of the 2nd kind from the integration path is equivalent to the condition that such an integral over any closed contour is equal to zero.

Ticket No. 34.Surface integral of the first kind (over surface area). Applications (mass of a material surface, coordinates of the center of gravity, moments, area of ​​a curved surface).

Consider an open surface S, limited by contour L, and divide it into parts by some curves S 1, S 2,…, S n. Let's select a point in each part M i and project this part onto a tangent plane to the surface passing through this point. We obtain in projection a flat figure with area T i. Let us call ρ the greatest distance between two points on any part of the surface S.

Definition 12.1. Let's call area S surfaces area sum limit T i at

Surface integral of the first kind.

Consider some surface S, limited by contour L, and break it into parts S 1, S 2,…, S p(we will also denote the area of ​​each part S p). Let the value of the function be specified at each point of this surface f(x, y, z). Let's choose in each part S i point M i (x i , y i , z i) and compose the integral sum

. (12.2)

Definition 12.2. If there is a finite limit for the integral sum (12.2), independent of the method of dividing the surface into parts and the choice of points M i, then it is called surface integral of the first kind from the function f(M) = f(x, y, z) on the surface S and is designated

Comment. A surface integral of the 1st kind has the usual properties of integrals (linearity, summation of integrals of a given function over individual parts of the surface under consideration, etc.).

Geometric and physical meaning of a surface integral of the 1st kind.

If the integrand f(M)≡ 1, then from Definition 12.2 it follows that it is equal to the area of ​​the surface under consideration S.

. (12.4)

Application of a surface integral of the 1st kind.

1. The area of ​​a curved surface, the equation of which is z = f(x, y), can be found in the form:

(14.21)

(Ω – projection S to the O plane xy).

2. Surface mass

(14.22)

3. Moments:

Static moments of the surface relative to the coordinate planes O xy, O xz, O yz;

Moments of inertia of the surface relative to the coordinate axes;

Moments of inertia of the surface relative to coordinate planes;

- (14.26)

The moment of inertia of the surface relative to the origin.

4. Coordinates of the surface center of mass:

. (14.27)

Ticket number 35. Calculation of the surface integral of the 1st kind (reducing it to a multiple).

Let us restrict ourselves to the case when the surface S is given explicitly, that is, by an equation of the form z = φ(x, y). Moreover, from the definition of surface area it follows that

S i =, where Δ σi – projection area S i to the O plane xy, A γ i– angle between the O axis z and normal to the surface S at the point M i. It is known that

,

Where ( x i , y i , z i) – point coordinates M i. Therefore,

Substituting this expression into formula (12.2), we obtain that

,

Where the summation on the right is carried out over the region Ω of the O plane xy, which is the projection onto this surface plane S(Fig. 1).

S: z=φ(x,y)

Δσ iΩ

In this case, on the right side, an integral sum is obtained for a function of two variables over a flat region, which in the limit at gives a double integral. Thus, a formula has been obtained that allows us to reduce the calculation of a surface integral of the 1st kind to the calculation of a double integral:

Comment. Let us clarify once again that on the left side of formula (12.5) there is surface integral, and on the right - double.

Ticket number 36.Surface integral of the second kind. Vector field flow. Relationship between surface integrals of the first and second kind.

Vector field flow.

Consider the vector field A (M), defined in the spatial domain G, oriented smooth surface S G and the field of unit normals P (M) on the selected side of the surface S.

Definition 13.3. Surface integral of the 1st kind

, (13.1)

Where An is the scalar product of the corresponding vectors, and A p– vector projection A to the normal direction is called vector field flow A(M) through the selected side of the surface S .

Remark 1. If you choose the other side of the surface, then the normal, and, consequently, the flux will change sign.

Remark 2. If the vector A specifies the speed of fluid flow at a given point, then integral (13.1) determines the amount of fluid flowing per unit time through the surface S in a positive direction (hence the common term "flow").

Ostrogradsky-Green formula

This formula establishes a connection between the curvilinear integral over a closed contour C and the double integral over the area limited by this contour.

Definition 1. A region D is called a simple region if it can be divided into a finite number of regions of the first type and, independently of this, into a finite number of regions of the second type.

Theorem 1. Let functions P(x,y) and Q(x,y) be defined in a simple domain and be continuous along with their partial derivatives and

Then the formula holds

where C is the closed contour of area D.

This is the Ostrogradsky-Green formula.

Conditions for the independence of a curvilinear integral from the path of integration

Definition 1. A closed squarable region D is said to be simply connected if any closed curve l D can be continuously deformed into a point so that all points of this curve would belong to the region D (region without “holes” - D 1), if such deformation is impossible, then the region is called multiply connected (with “holes” - D 2).

Definition 2. If the value of a curve integral along a curve AB does not depend on the type of curve connecting points A and B, then this curve integral is said to be independent of the path of integration:

Theorem 1. Let the continuous functions P(x,y) and Q(x,y) be defined in a closed simply connected domain D, together with their partial derivatives. Then the following 4 conditions are equivalent:

1) curvilinear integral over a closed loop

where C is any closed loop in D;

2) the curvilinear integral over a closed loop does not depend on the path of integration in region D, i.e.

3) the differential form P(x,y)dx + Q(x,y)dy is the total differential of some function F in the domain D, i.e., there is a function F such that (x,y) D the equality holds

dF(x,y) = P(x,y)dx + Q(x,y)dy; (3)

4) for all points (x,y) D the following condition will be satisfied:

Let's prove it using the diagram.

Let's prove that from.

Let 1 be given), i.e. = 0 by property 2 §1, which = 0 (by property 1 §1) .

Let's prove that from.

It is given that cr.int. does not depend on the path of integration, but only on the choice of the beginning and end of the path

Consider the function

Let us show that the differential form P(x,y)dx + Q(x,y)dy is the complete differential of the function F(x,y), i.e. , What

Let's set the private growth

x F (x,y)= F(x + x, y) -F (x,y)= = == =

(by property 3 of § 1, BB* Oy) = = P (c,y)x (by the mean value theorem, c -const), where x

(due to the continuity of the function P). We obtained formula (5). Formula (6) is obtained similarly.

Let's prove that from.

The formula is given

dF(x,y) = P(x,y)dx + Q(x,y)dy.

Obviously = P(x,y). Then

According to the conditions of the theorem, the right-hand sides of equalities (7) and (8) are continuous functions, then, by the theorem on the equality of mixed derivatives, the left-hand sides will also be equal, i.e., that

Let us prove that out of 41.

Let us choose any closed contour from the region D that bounds the region D 1 .

The functions P and Q satisfy the Ostrogradsky-Green conditions:

By virtue of equality (4), on the left side of (9) the integral is equal to 0, which means that the right side of the equality is also equal to

Remark 1. Theorem 1 can be formulated in the form of three independent theorems

Theorem 1*. In order for a simply connected squarable domain D to have a curved int. did not depend on the path of integration so that condition (.1) is satisfied, i.e.

Theorem 2*. In order for a simply connected squarable domain D to have a curved int. did not depend on the path of integration so that condition (3) is satisfied:

the differential form P(x,y)dx + Q(x,y)dy is the total differential of some function F in the domain D.

Theorem 3*. In order for a simply connected squarable domain D to have a curved int. did not depend on the path of integration so that condition (4) is satisfied:

Remark 2. In Theorem 2*, the domain D can also be multiply connected.

2nd kind from the path of integration

Consider a curvilinear integral of the 2nd kind, where L is the curve connecting the points M and N. Let the functions P(x, y) and Q(x, y) have continuous partial derivatives in some domain D in which the curve L lies entirely. Let us determine the conditions under which the curvilinear integral under consideration does not depend on the shape of the curve L, but only on the location of the points M and N.

Let's draw two arbitrary curves MSN and MTN, lying in area D and connecting points M and N (Fig. 14).

Let us assume that, that is,

where L is a closed loop made up of the MSN and NTM curves (hence, it can be considered arbitrary). Thus, the condition for the independence of a curvilinear integral of the 2nd kind from the integration path is equivalent to the condition that such an integral over any closed contour is equal to zero.

Theorem 5 (Green's theorem). Let the functions P(x, y) and Q(x, y) and their partial derivatives and be continuous at all points of some domain D. Then, in order for any closed contour L lying in the domain D to satisfy the condition

it is necessary and sufficient that = at all points of the region D.

Proof.

1) Sufficiency: let the condition = be satisfied. Let us consider an arbitrary closed contour L in the region D, bounding the region S, and write Green’s formula for it:

So, sufficiency has been proven.

2) Necessity: suppose that the condition is satisfied at every point of the region D, but there is at least one point of this region at which -? 0. Let, for example, at the point P(x0, y0) we have: - > 0. Since the left side of the inequality contains a continuous function, will it be positive and greater than some? > 0 in some small region D` containing the point P. Consequently,

From here, using Green’s formula, we obtain that

where L` is the contour limiting the area D`. This result contradicts the condition. Consequently, = at all points of the region D, which is what needed to be proved.

Remark 1. Similarly, for three-dimensional space it can be proven that the necessary and sufficient conditions for the independence of the curvilinear integral

from the integration path are:

Remark 2. If conditions (52) are met, the expression Pdx + Qdy + Rdz is the total differential of some function u. This allows us to reduce the calculation of a curvilinear integral to determining the difference between the values ​​at both the final and initial points of the integration contour, since

In this case, the function and can be found using the formula

where (x0, y0, z0) is a point from region D, and C is an arbitrary constant. Indeed, it is easy to verify that the partial derivatives of the function and, given by formula (53), are equal to P, Q and R.

Example 10.

Calculate line integral of the 2nd kind

along an arbitrary curve connecting points (1, 1, 1) and (2, 3, 4).

Let us make sure that conditions (52) are met:

Therefore, the function exists. Let us find it using formula (53), putting x0 = y0 = z0 = 0. Then

Thus, the function is determined up to an arbitrary constant term. Let's take C = 0, then u = xyz. Hence,