These formulas relate an integral over a figure to some integral over the boundary of the given figure.

Let the functions be continuous in the domain DÌ Oxy and on its border G; region D- connected; G is a piecewise smooth curve. Then true Green's formula:

here on the left is a curvilinear integral of the first kind, on the right is a double integral; circuit G turns counterclockwise.

Let T is a piecewise-smooth bounded two-sided surface with a piecewise-smooth boundary G. If functions P(x,y,z), Q(x,y,z), R(x,y,z) and their first-order partial derivatives are continuous at points of the surface T and borders G, then Stokes formula:

(2.23)

on the left is a curvilinear integral of the second kind; on the right - the surface integral of the second kind, taken on the other side of the surface T, which remains on the left when traversing the curve G.

If the connected area WÌ Oxyz bounded by a piecewise-smooth, closed surface T, and the functions P(x,y,z), Q(x,y,z), R(x,y,z) and their first-order partial derivatives are continuous at points from W And T, then Ostrogradsky-Gauss formula:

(2.24)

on the left, a surface integral of the second kind over the outer side of the surface T; on the right is the triple integral over the area W.

Example 1 Calculate the work of force when going around the point of its application of the circle G: , starting from the axis Ox, clockwise (Fig. 2.18).

Solution. Work equals . Let us apply Green's formula (2.22), putting the “-” sign on the right in front of the integral (since the circuit is traversed clockwise) and taking into account that P(x,y)=x-y, Q(x,y)=x+y. We have:
,
Where S D- area of ​​a circle D: , equal to . As a result: - the desired work of the force.

Example 2 Calculate Integral , If G there is a circle in the plane z=2, bypassed counterclockwise.

Solution. By the Stokes formula (2.23), we reduce the original integral to the surface integral over the circle T:
T:

So, considering that , we have:

The last integral is the double circle integral DÌ Oxy, on which the circle was projected T; D: . Let's move on to polar coordinates: x=r cosj, y=r sinj, jО, r O. Eventually:
.

Example 3 Find a flow P T pyramids W: (Fig. 2.19) in the direction of the outer normal to the surface.

Solution. The flow is . Applying the Ostrogradsky-Gauss formula (2.24), we reduce the problem to the calculation of the triple integral over the figure W-pyramid:

Example 4 Find a flow P vector field through the full surface T pyramids W: ; (Fig. 2.20), in the direction of the outer normal to the surface.

Solution. We apply the Ostrogradsky-Gauss formula (2.24) , where V is the volume of the pyramid. Let's compare with the solution of direct calculation of the flow ( – pyramid faces).

,
since the projection of the faces onto the plane Oxy has zero area (Fig. 2.21),

(Ostrogradsky Mikhail Vasilievich (1861–1862) - Russian mathematician,

Academician Petersburg. A.N.)

(George Green (1793 - 1841) - English mathematician)

Sometimes this formula is called Green's formula, however, J.

Green proposed in 1828 only a special case of the formula.

The Ostrogradsky-Green formula establishes a connection between the curvilinear integral and the double integral, i.e. gives an expression for an integral over a closed contour in terms of a double integral over the region bounded by this contour.

If the closed loop has the form shown in the figure, then the curvilinear integral over the loop L can be written as:

If sections AB and CD of the contour are taken as arbitrary curves, then, after carrying out similar transformations, we obtain a formula for an arbitrary-shaped contour:

This formula is called the Ostrogradsky-Green formula.

The Ostrogradsky-Green formula is also valid in the case of a multiply connected domain, i.e. area within which there are excluded areas. In this case, the right side of the formula will be the sum of the integrals over the outer contour of the region and the integrals over the contours of all excluded sections, and each of these contours is integrated in such a direction that the region D always remains on the left side of the bypass line.

Example. Let's solve the example discussed above using the Ostrogradsky-Green formula.

The Ostrogradsky-Green formula makes it possible to significantly simplify the calculation of the curvilinear integral.

The curvilinear integral does not depend on the shape of the path if it has the same value along all paths connecting the starting and ending points.

The condition of the independence of the curvilinear integral from the shape of the path is equivalent to the equality of this integral to zero over any closed contour containing the start and end points.

This condition will be satisfied if the integrand is the total differential of some function, i.e. the totality condition is satisfied.

(Ostrogradsky Mikhail Vasilyevich (1861-1862) - Russian mathematician,

Academician Petersburg. A.N.)

(George Green (1793 - 1841) - English mathematician)

Sometimes this formula is called Green's formula, however, J. Green proposed in 1828 only a special case of the formula.

The Ostrogradsky-Green formula establishes a connection between the curvilinear integral and the double integral, i.e., it gives an expression for the integral over a closed contour in terms of a double integral over the region bounded by this contour.

If the closed loop has the form shown in the figure, then the curvilinear integral over the loop L can be written as:

If sections AB and CD of the contour are taken as arbitrary curves, then, after carrying out similar transformations, we obtain a formula for an arbitrary-shaped contour:

This formula is called Ostrogradsky-Green formula.

The Ostrogradsky-Green formula is also valid in the case of a multiply connected region, i.e., a region within which there are excluded sections. In this case, the right side of the formula will be the sum of the integrals over the outer contour of the region and the integrals over the contours of all excluded sections, and each of these contours is integrated in such a direction that the region D always remains on the left side of the bypass line.

Example. Let's solve the example discussed above using the Ostrogradsky-Green formula.

The Ostrogradsky-Green formula makes it possible to significantly simplify the calculation of the curvilinear integral.

The curvilinear integral does not depend on the shape of the path if it has the same value along all paths connecting the starting and ending points.

The condition of the independence of the curvilinear integral from the shape of the path is equivalent to the equality of this integral to zero over any closed contour containing the start and end points.

Communication between dv. Int. In area D and curviline. Int. The Ostrogradsky-Green formula is established for the region L.

Let the domain D limit be given on the plane OXY. Curve intersecting with straight parallel cords. Axes at no more than 2 points, i.e., the area D is correct.

T1.If f. P(x,y), Q(x,y) is continuous with its multiple derivatives ,

The area D is then fair forms. (f.Ostr.-Gr.)

L is the boundary of the region D and integration along the curve L is performed in the positive direction.Dovo.

T2. If = (2), then the subintegr. The expression P*dx+Q*dy yavl. Full diff. Functions U=U(x,y).

P*dx+Q*dy =U(x.y)

Satisfies condition (2) can be found using f.

Note 1 In order not to confuse the variable integr. X with the upper prednl its designator. Another letter.

Deputy 2 as the start point (x0,Y0) usually take the point (0.0)

Condition of independence of curvilinear int. 2nd kind from the path integr.

Let t. A (X1, Y1), B(X2, Y2),. Let the product points of area D. Points A and B can be connected by different lines. For each of them, Int. will have its own value, if the value is the same for all curves, then the integral does not depend on the type of path int., in this case it is enough to note the initial. Point A (X1, Y1) and endpoint B(X2, Y2).

T. In order for cr. Int.

Does not depend on the path int. Area D in the cat. F. P(X,Y), Q(X,Y) are continuous together with their derivatives and it is necessary that at each point of the region = Dok-in

Cr. Int. 2nd kind does not depend on the path of integration

Deputy = hence we get that

Pov. Int. 1st kind. His St. and calc.

Let at the points pov. S C PL. S space oxyz def. continuous f. f(x,y.z) .

Let's break pov. S into n parts Si, PL. EACH PART delta Si, and the diameter Di i=1..m in each part Si choose an arbitrary point Mi from (xi, yi, zi) and make the sum . The sum is called integral for f. f(x,y.z) over the surface S if for integral. The sum has a limit, it is called. Pov integral of the 1st kind from f. f(x,y.z) over the surface S and denoted by =

Surface properties Int.

2) 3) S=s1+s2, Then 4) f1<=f2 , т о 5) 6) 7) Ф. f непрерывна на поверхности S , то на этой поверхности сущ. Точка M(x0,y0,z0) S, такая, что .

Calculation of int of the 1st kind be reduced to the calculation of the 2nd int according to region D, which is the projection of the ep S onto the oxy plane, if the ep s is given by Ur z=z(x, y), then the ep is equal to .

If S is given as y=y(x, z), then...

Pov int 2nd kind

Let a two-sided surface be given, after going around such a surface without crossing its boundary, the direction of the normal to it does not change. One-sided pov: is a Möbius strip. Let φ be defined at a point of the considered two-sided surface S in the space oxyz. F(x,y,z). We divide the ejected side of the surface into parts Si i=1..m and project them onto the cord of the plane. In this case, pl pov, we take with the “+” sign if the upper side of the pov is selected (if the normal forms an acute angle with oz, select with the “-” sign if the lower side of the pov is selected (OBTE ANGLE)). Let us compose an int sum Where - pl pov Si -parts for if it exists and does not depend on the method of dividing the surface into parts and on the choice of points in them, we call int of the 2nd kind from f. f(x,y,z) over s and is denoted: by definition, the integral will be = the limit of the sum integral. Similarly, int over s

, then the general view of the int of the 2nd kind is int where P, Q, R are continuous functions defined at the points of the two-sided surface s. If S is closed pov, then int on the outside is denoted and on the inside. ds. Where ds is the area element of S , and cos , cos cos for example cos is n. The selected side of the pov.

Relationship between double integral over area D and the curvilinear integral over the boundary L This area is established by the Ostrogradsky-Green formula, which is widely used in mathematical analysis.

Let on the plane Ohu area is set D, bounded by a curve intersecting with straight lines parallel to the coordinate axes at no more than two points, i.e. region D- correct.

Theorem 10.2. If functions P(x; y) And Q(x; y) are continuous together with their partial derivatives And in area D, then the formula

(10.8)

Where L– area border D and integration along the curve L produced in the positive direction (i.e. when moving along the curve, the area D stays on the left).

Formula (10.8) is called the Ostrogradsky-Green formula.

P mouth
- arc equation AnB, A
- arc equation AmB(see fig. 8). Let's find first
.According to the rule for calculating the double integral, we have:

Or according to formula (10.6), Fig. 8.

Similarly, it is proved that
(10.10)

If we subtract equality (10.9) from equality (10.10), then we obtain formula (10.8).

Comment. Formula (10.8) is also valid for an arbitrary region, which can be divided into a finite number of regular regions.

Example 10.3. Using the Ostrogradsky-Green formula, calculate

Where L- contour of a rectangle with vertices A(3;2 ), IN(6;2 ), WITH(6;4 ), D(3;4 ).

○ Solution: Figure 9 shows the integration contour. Because the

by formula (10.8) we have:

10.4. Conditions for the independence of a curvilinear integral of the second kind from the path of integration

P
mouth A(x 1 ; y 1) and B(x 2 ; y 2) are two arbitrary points of a simply connected area D plane Ohu(plane D called singly connected , if for any closed contour lying in this region, the part of the plane bounded by it belongs entirely to D(area without "holes")). points A And IN can be connected by various lines (in Fig. 10 this L 1 , L 2 and L 3). Over each of these curves, the integral
has, in general, its own meaning.

If its values ​​for all possible curves AB are the same, then we say that the integral I does not depend on the type of integration path.

Rice. 10. In this case, for the integral I it suffices to note only its starting point A(x 1 ; y 1 ) and its endpoint B(x 2 ; y 2 ) way. Write down:

(10.11)

What are the conditions under which the curvilinear integral of the second kind does not depend on the form of the integration path?

Theorem 10.3. In order for the curvilinear integral
did not depend on the integration path in a simply connected domain D, in which the functions P(x; y), Q(x; y) are continuous together with their partial derivatives, it is necessary and sufficient that the condition =(10.12)

Let us prove the sufficiency of condition (10.12). Consider an arbitrary closed circle AmBnA(or L) in area D(see fig. 11). It satisfies the Ostrogradsky-Green formula (10.8) By virtue of condition (10.12), we have:
, or
. Taking into account the properties of the curvilinear integral, we have:

, i.e.

The resulting equality means that the curvilinear integral does not depend on the path of integration.

Fig.11. In the course of the proof of the theorem, it was obtained that if the condition =, then the integral over a closed circle is equal to zero:

The converse is also true.

Corollary 10.1. If condition (10.12) is satisfied, then the integrand is the total differential of some function u = u(x; y), i.e.

Then (see (10.11))

Formula (10.14) is called the generalized Newton-Leibniz formula for the curvilinear integral of the total differential.

Corollary 10.2. If the integrand pdx + Qdy there is a total differential and an integration path L closed, then
.

Notes:

As a starting point ( x 0 ; y 0) usually take the point (0;0) - the origin (see example 10.5).

= ,=,=;

Example 10.4. Find

Solution: Here P = y, Q = x, == 1. According to the above theorem, the integral does not depend on the path of integration. As an integration path, you can take a straight line segment y = x, the arc of the parabola y = x 2 etc. or use formula (10.14). Because ydx + xdy = d(xy), That

Example 10.5. Make sure the expression is the full differential of the function U (x; y) and find it.

Solution: In order for the indicated expression to be a total differential, conditions (10.12) must be satisfied:

The conditions are satisfied, therefore, And since the total differential has the form

,

then the relations

(10.16)

Integrating over X the first of the equations, assuming at constant, while instead of constantly integrating one should put
is an unknown function that depends only on at:

Substituting the resulting expression into the second equation (10.16), we find
:

Thus,

Note that the function U easier to find using formula (10.15).