Conditions for the independence of the curvilinear integral from the path of integration on the plane. Conditions for independence of a curvilinear integral of the second kind from the path of integration Independence of a curvilinear integral from a contour
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taken along some plane curve L connecting the 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.
Consider two arbitrary curves MPN and MQN, lying in the region D under consideration and connecting the points M and N (Fig. 351). Let
Then, on the basis of properties 1 and 2 of the curvilinear integrals (§ 1), we have
i.e. closed-loop curvilinear integral
In the last formula, the curvilinear 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 curvilinear 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 curvilinear integral over any closed contour is equal to zero.
The converse conclusion is also true: if the 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 integration path, in example 3, the curvilinear integral depends on the integration path, since in this example the integral over a closed contour does not equal zero, but gives the area bounded by the contour under consideration; in examples 1 and 2, the curvilinear integrals also depend on the integration path.
The question naturally arises: what conditions must the functions satisfy in order for the curvilinear integral over 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, and be continuous at all points of some domain D. Then, in order for the curvilinear integral over any closed contour L lying in the domain D to be equal to zero, i.e., so that
it is necessary and sufficient that the equality
in all estrus areas
Proof. 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 proved.
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 region D, then condition (3) is also satisfied at each point of this region.
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 there is a continuous function on the left side of the inequality, it will be positive and greater than some number at all points of some sufficiently small area D containing the point . Let us take the double integral over this region of the difference . It will be positive. Really,
But according to Green's formula, the left side of the last inequality is equal to the curvilinear integral over 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 false. From here
it follows that
at all points in the area
Thus, the theorem is completely proved.
In § 9 ch. XIII, it was proved 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
is the 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 the points M and N, (M) is equal to the difference between the values of the function and at these points:
Proof. If is the total differential of the function, then the curvilinear integral takes the form
To calculate this integral, we write the parametric equations of the curve L connecting the points M and
integral, reduces to the following definite integral:
The expression in brackets is a function of which is the total derivative of the function. Therefore
As we can see, the curvilinear integral of the total differential does not depend on the shape of the curve along which the integration is performed.
A similar assertion also holds for a curvilinear integral over a space curve (see § 7 below).
Comment. Sometimes it is necessary to consider curvilinear integrals over the arc length L of some function
30. Green's formula. Conditions for the independence of the curvilinear integral from the path of integration.
Green's formula: If C is the closed boundary of the domain D and the functions P(x,y) and Q(x,y), together with their first-order partial derivatives, are continuous in the closed domain D (including the boundary of C), then Green's formula is valid:, and the bypass around contour C is chosen so that region D remains on the left.
From the lectures: Let the functions P(x,y) and Q(x,y) be given, which are continuous in the domain D together with partial derivatives of the first order. The boundary integral (L) lying entirely in the region D and containing all points in the region D: . The positive direction of the contour is when the limited part of the contour is on the left.
Condition of independence of the curvilinear integral of the 2nd kind from the path of integration. A necessary and sufficient condition for the fact that the curvilinear integral of the first kind, connecting the points M1 and M2, does not depend on the path of integration, but depends only on the initial and final points, is the equality:.
.
31. Surface integrals of the first and second kind, their main properties and calculation.
- specifying the surface.
We project S onto the xy plane, we get the area D. We divide the area D with a grid of lines into parts called Di. From each point of each line we draw lines parallel to z, then S will be divided into Si. Let's make an integral sum: . Let us set the maximum diameter Di to zero:, we get:
This is a surface integral of the first kind
This is the surface integral of the first kind.
Definition in brief. If there is a finite limit of the integral sum, which does not depend on the method of partitioning S into elementary sections Si and on the choice of points, then it is called a surface integral of the first kind.
When passing from variables x and y to u and v:
P 
a surface integral has all the properties of an ordinary integral. See questions above.
Definition of a surface integral of the second kind, its main properties and calculation. Connection with the integral of the first kind.
Let a surface S bounded by a line L (Fig. 3.10) be given. Take some contour L on the surface S that does not have common points with the boundary L. At the point M of the contour L, two normals u can be restored to the surface S. We choose one of these directions. Outline the point M along the contour L with the selected direction of the normal.
If the point M returns to its original position with the same direction of the normal (and not with the opposite direction), then the surface S is called two-sided. We will consider only two-sided surfaces. A two-sided surface is any smooth surface with the equation .
Let S be a two-sided non-closed surface bounded by a line L that has no self-intersection points. Let's choose a certain side of the surface. We will call the positive direction of bypassing the contour L such a direction, when moving along which along the selected side of the surface, the surface itself remains on the left. A two-sided surface with a positive direction of contour traversal set on it in this way is called an oriented surface.
Let us proceed to the construction of a surface integral of the second kind. Let us take a two-sided surface S in space, consisting of a finite number of pieces, each of which is given by an equation of the form or is a cylindrical surface with generators parallel to the Oz axis.
Let R(x,y,z) be a function that is defined and continuous on the surface S. Using a network of lines, we partition S arbitrarily into n "elementary" segments ΔS1, ΔS2, ..., ΔSi, ..., ΔSn that do not have common internal points. On each segment ΔSi, we arbitrarily choose a point Mi(xi,yi,zi) (i=1,...,n). Let (ΔSi)xy be the projection area of the section ΔSi onto the Oxy coordinate plane, taken with the "+" sign, if the normal to the surface S at the point Mi(xi,yi,zi) (i=1,...,n) forms with axis Oz is an acute angle, and with the sign "-" if this angle is obtuse. Let us compose the integral sum for the function R(x,y,z) over the surface S with respect to the variables x,y: . Let λ be the largest of the diameters ΔSi (i = 1, ..., n).
If there is a finite limit that does not depend on the method of partitioning the surface S into "elementary" sections ΔSi and on the choice of points, then it is called the surface integral over the selected side of the surface S of the function R (x, y, z) along the coordinates x, y (or surface integral of the second kind) and is denoted 
.
Similarly, one can construct surface integrals over the coordinates x, z or y, z along the corresponding side of the surface, i.e. 
And 
.
If all these integrals exist, then you can introduce a "general" integral over the selected side of the surface: .
A surface integral of the second kind has the usual properties of an integral. We only note that any surface integral of the second kind changes sign when the side of the surface changes.
Connection between surface integrals of the first and second kind.
Let the surface S be given by the equation: z \u003d f (x, y), and f (x, y), f "x (x, y), f "y (x, y) are continuous functions in a closed region τ (projections of the surface S onto the Oxy coordinate plane), and the function R(x,y,z) is continuous on the surface S. The normal to the surface S, which has direction cosines cos α, cos β, cos γ, is chosen to the upper side of the surface S. Then .
For the general case, we have:
= 
| " |
Consider a curvilinear integral of the 2nd kind , where L- 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 considered curvilinear integral does not depend on the shape of the curve L, but only on the location of the points M And N.
Draw two arbitrary curves MPN And MQN, lying in the area D and connecting points M And N(Fig. 1).
M N Rice. 1. P
Let's assume that , that is
Then where L- a closed contour, composed of curves MPN And NQM(hence, it can be considered arbitrary). Thus, the condition for the independence of a curvilinear integral of the 2nd kind from the path of integration is equivalent to the condition that such an integral over any closed contour is equal to zero.
Theorem 1. Let at all points of some area D functions are continuous P(x, y) And Q(x, y) and their partial derivatives and . Then in order for any closed loop L, lying in the region D, the condition
It is necessary and sufficient that = at all points of the region D.
Proof .
1) Sufficiency: let the condition = be fulfilled. Consider an arbitrary closed loop L in area D, limiting the area S, and write Green's formula for it:
So, sufficiency is proved.
2) Necessity: suppose the condition is met at every point of the area D, but there is at least one point in this region where - ≠ 0. Let, for example, at the point P(x0, y0)-> 0. Since there is a continuous function on the left side of the inequality, it will be positive and greater than some δ > 0 in some small area D` containing point R. Hence,
Hence, by Green's formula, we obtain that , where L`- the outline that bounds the area D`. This result contradicts the condition . Therefore, = at all points of the region D, which was to be proved.
Remark 1 . Similarly, for a three-dimensional space, one can prove that the necessary and sufficient conditions for the independence of the curvilinear integral
from the path of integration are:
Remark 2. When the conditions (28/1.18) are met, the expression Pdx+Qdy+Rdz is the total differential of some function And. This allows us to reduce the calculation of the curvilinear integral to the determination of the difference between the values And at the end and start points of the integration contour, since
At the same time, the function And can be found using the formula
Where ( x0, y0, z0)– point from area D, a C is an arbitrary constant. Indeed, it is easy to verify that the partial derivatives of the functions And given by formula (28/1.19) are P, Q And R.
Ostrogradsky-Green formula
This formula establishes a connection between the curvilinear integral over a closed contour C and the double integral over the area bounded by this contour.
Definition 1. A domain D is called a simple domain if it can be partitioned into a finite number of domains of the first type and, independently of this, into a finite number of domains of the second type.
Theorem 1. Let the functions P(x,y) and Q(x,y) be defined in a simple domain, which are continuous together with their partial derivatives and
Then the formula holds
where С is a closed contour of the region D.
This is the Ostrogradsky-Green formula.
Conditions for the independence of the 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 (a 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 the curvilinear integral along the curve AB does not depend on the type of the curve connecting the points A and B, then they say that this curvilinear integral does not depend on the integration path:
Theorem 1. Let in a closed simply connected domain D be defined continuous functions P(x,y) and Q(x,y) together with their partial derivatives. Then the following 4 conditions are equivalent (equivalent):
1) curvilinear integral along a closed contour
where C is any closed loop in D;
2) the curvilinear integral over a closed contour does not depend on the path of integration in the domain 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., that there exists a function F such that (x,y)D the equality
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 according to the scheme.
Let us prove that from
Let 1), i.e., be given = 0 by property 2 of §1, which = 0 (by property 1 of §1) .
Let us 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 total differential of the function F(x,y), i.e., , What
Let's set a private gain
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, with -const), where x (due to the continuity of the function P). We have obtained formula (5). Formula (6) is obtained similarly. Let us prove that from Given the formula dF(x,y) = P(x,y)dx + Q(x,y)dy. Obviously, = P(x, y). Then By the condition of the theorem, the right parts of equalities (7) and (8) are continuous functions, then by the theorem on the equality of mixed derivatives, the left parts will also be equal, i.e., that Let us prove that out of 41. Let's choose any closed contour from the region D, which limits 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 equal to Remark 1. Theorem 1. can be formulated as three independent theorems Theorem 1*. In order for the curved int. does not depend on the integration path so that condition (.1) is satisfied, i.e. Theorem 2*. In order for the curved int. does not depend on the integration path 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 the curved int. does not depend on the integration path so that condition (4) is satisfied: Remark 2. In Theorem 2*, the domain D can also be multiply connected. Definition. The region G of three-dimensional space is called surface simply connected. if any closed contour lying in this region can be spanned by a surface that lies entirely in the region G. For example, the interior of a sphere or the entire three-dimensional space are surface simply connected regions; the interior of a torus or a three-dimensional space, from which the line is excluded, are not surface simply connected regions. Let a continuous vector field be given in a surface simply connected domain G. Then the following theorem holds. Theorem 9. In order for the curvilinear integral in the field of the vector a not to depend on the integration path, but to depend only on the initial and final points of the path (A and B), it is necessary and sufficient that the circulation of the vector a along any closed contour L located in the region G, was equal to zero. 4 Necessity. Let the m-integral also be independent of the integration path. Let us show that then for any closed contour L is equal to zero. Consider an arbitrary closed contour L in the field of the vector a and take arbitrary points A and B on it (Fig. 35). By condition, we have - different paths connecting exactly A and B \ from where exactly the chosen closed contour L is. Sufficiency. Let L be for any closed contour. Let us show that in this case the integral does not depend on the integration path. Let us take two points A and B in the field of the vector a, connect them by arbitrary lines L1 and L2, and show that For simplicity, we confine ourselves to the case when the lines L1 and L2 do not intersect. In this case, the union forms a simple closed loop L (Fig. 36). By condition a, by the property of additivity. Independence of the curvilinear integral from the path of integration Potential field Calculation of the curvilinear integral in the potential field Calculation of the potential in Cartesian coordinates Hence, from which the validity of equality (2) follows. Theorem 9 expresses the necessary and sufficient conditions for the independence of the curvilinear integral from the shape of the path, but these conditions are difficult to verify. Let us present a more efficient criterion. Theorem 10. For the curvilinear integral to be independent of the integration path L, it is necessary and sufficient that the vector field be irrotational. M) is superficially simply connected. Comment. By virtue of Theorem 9, the independence of the curvilinear integral from the path of integration is equivalent to the equality to zero of the circulation of the vector a along any closed contour. We use this circumstance in the proof of the theorem. Necessity. Let the curvilinear integral be independent of the shape of the path, or, what is the same, the circulation of the vector a along any closed contour L is equal to zero. Then, i.e., at each point of the field, the projection of the vector rot a on any direction is equal to zero. This means that the vector rot a itself is equal to zero at all points of the field, Sufficiency. The sufficiency of condition (3) follows from the Stokes formula, since if rot a = 0, then the circulation of the vector along any closed loop L is equal to zero: The rotor of a flat field is equal to which allows us to formulate the following theorem for a flat field. Theorem 11. In order for the curvilinear integral in a simply connected plane field to be independent of the shape of the line L, it is necessary and sufficient that the relation holds identically in the entire region under consideration. If the domain is not simply connected, then the fulfillment of the condition, generally speaking, does not ensure the independence of the curvilinear integral from the shape of the line. Example. Let Consider the integral It is clear that the integrand has no meaning at the point 0(0,0). Therefore, we exclude this point. In the rest of the plane (this will no longer be a simply connected region!) The coordinates of the vector a are continuous, have continuous partial derivatives and Let us consider the integral (6) along a closed curve L - a circle of radius R centered at the origin: Then The difference of the circulation from zero shows that integral (6) depends on the form of the integration path. §10. Potential field Definition. The field of the vector a(M) is called potential if there exists a scalar function u(M) such that the function u(M) is called the potential of the field; its level surfaces are called equipotential surfaces. then relation (1) is equivalent to the following three scalar equalities: Note that the field potential is determined up to a constant term: if, therefore, is a constant number. Example 1. The field of the radius vector r is potential, since we recall that the Potential of the field of the radius vector is, therefore. Example 2. The vector field is potential. Let the function be such that it is found. Then and from where Hence, is the potential of the field. Theorem 12. In order for the vector a to be potential, it is necessary and sufficient that it be irrotational, i.e., that its rotor be equal to zero at all points of the field. In this case, the continuity of all partial derivatives of the coordinates of the vector a and the surface simply connectedness of the region in which the vector a is given are assumed. Necessity. The necessity of condition (2) is established by direct calculation: if the field is potential, i.e., by virtue of the independence of the mixed derivatives from the order of differentiation. Adequacy. Let the vector field be irrotational (2). In order to prove the potentiality of this field, we construct its potential u(M). It follows from condition (2) that the curvilinear integral does not depend on the shape of the line L, but depends only on its initial and final points. Let us fix the starting point and change the end point Mu, z). Then the integral (3) will be a function of the point. Let us denote this function by u(M) and prove that In what follows, we will write the integral (3), indicating only the start and end points of the integration path, Equality is equivalent to three scalar equalities Independence of the curvilinear integral from the integration path Potential field Potential in Cartesian Coordinates Let us prove the first of them; the second and third equalities are proved in a similar way. By definition of a partial derivative, we have Consider a point close to a point Since the function u(M) is determined by relation (4), in which the curvilinear integral does not depend on the integration path, we choose the integration path as indicated in Fig.37. Then From here the last integral is taken as a mole of a segment of the straight line MM) parallel to the Ox axis. On this segment, the coordinate x can be taken as a parameter: Applying the mean value theorem to the integral on the right side of (6), we obtain From formula (7) it follows that Since then, due to the continuity of the function, we obtain Similarly, it is proved that Corollary. A vector field is potential if and only if the curvilinear integral in it is path-independent. Calculation of a curvilinear integral in a potential field Theorem 13. The integral in a potential field a(M) is equal to the difference between the values of the potential u(M) of the field at the end and initial points of the integration path. Previously, it was proved that the function is the potential of the field. In a potential field, a curvilinear intefal does not depend on the intefation point. Therefore, choosing the path of the point M\ to the point M2 so that it passes through the point Afo (Fig. 38), we obtain or, changing the orientation of the path in the first intefal on the right, Since the field potential is determined to within a constant term, then any potential of the considered fields can be written as where c is a constant. Making the substitution u - c in formula (10), we obtain the required formula for an arbitrary potential v(M) Example 3. In example 1, it was shown that the potential of the field of the radius vector r is the function where is the distance from the point to the origin. Calculation of the potential in Cartesian coordinates Let the potential field be given Previously, it was shown that the potential function "(M) can be found by the formula Integral (11) is most conveniently calculated as follows: we fix the starting point and connect it with a fairly close current point M(x, y ,z) a broken line whose links are parallel to the coordinate axes, . In this case, only one coordinate changes on each link of the polyline, which makes it possible to significantly simplify the calculations. Indeed, on the segment M0M\ we have: On the segment. Rice. 39. On the cut. Therefore, the potential is where are the coordinates of the current point on the links of the polyline, along which the integration is carried out. Example 4. Prove that the vector field k is potential and find its potential. 4 Let's check whether the field of the vector a(Af) is potential. With this goal, we calculate the rotor of the field. We have The field is potential. We find the potential of this field using formula (12). Let us take the origin of coordinates 0 as the initial point A/o (this is usually done if the field a(M) is defined at the origin of coordinates). Then we get So, where c is an arbitrary constant. The potential of this field can be found in another way. By definition, the potential u(x, y, z) is a scalar function for which gradu = a. This vector equality is equivalent to three scalar equalities: Integrating (13) over x, we obtain (17) by y, we find - some function z. Substituting (18) into (16), we get. Differentiating the last equality no z and taking into account relation (15), we obtain an equation for whence
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