2nd kind of way integration

Let's look at the curvilinear integral of the 2nd kind, where L is a curve that connects the points M and N. Let the functions P(x, y) and Q(x, y) be able to uninterruptedly walk privately in the real area D, in which surface lie Curve L. Significantly, in some analyses, the curvilinear integral cannot be deposited in the shape of the L curve, only the points M and N are expanded.

We draw two additional curves MSN and MTN, which lie in the distance D and connect points M and N (Fig. 14).

Let's say, what, tobto

de L - closed loop, folding from MSN and NTM curves (also, it can be added more). In this way, the mind's independence of the curvilinear integral of the 2nd kind in the way of integration is equal to the mind that such an integral behind any closed loop is equal to zero.

Theorem 5 (Green's theorem). Give at all points of the real area D the functions P(x, y) and Q(x, y) and their private transitions. Then, in order, for any closed contour L, which lies in the region D, the mind was

necessary and sufficient for all points of the region D.

Bringing.

1) Prosperity: let your mind go = vikonano. Let's look at a more closed contour L of the region D, which surrounds the region S, and write Green's formula for the new one:

Otzhe, sufficiency brought.

2) Necessity: let's say that the mind has been inscribed in the skin point of area D, but if you want to find one point in the center of the area, in which -? 0. Come on, for example, at the point P(x0, y0) maybe: - > 0. So, as the left part of the nervousness has an uninterrupted function, will it be positive and bigger for a day? > 0 in deakіy small region D`, to avenge the point P. Otzhe,

Consideration for Green's formula is essential that

de L` - the contour that surrounds the area D`. Tsey result superchit mind. Also, = at all points of the region D, which was necessary to bring.

Respect 1. Similarly, for a trivi-worldly space, one can bring that the necessary and sufficient minds of the independence of the curvilinear integral

in the direction of integration є:

Note 2. When vykonannі minds (52) viraz Pdx + Qdy + Rdz є top differential actual function. This allows you to calculate the calculation of the curvilinear integral up to the specified difference value and at the end and corn points of the contour of integration, scaling

For which function i can be known from the formula

de (x0, y0, z0) - the point of the region D, and C - has become enough. Indeed, it is easy to confuse that the private functions given by formula (53) are equal to P, Q and R.

butt 10.

Calculate the curvilinear integral of the 2nd kind

along the full curve, which connects the points (1, 1, 1) and (2, 3, 4).

Let's move on, what the vicons think (52):

Otzhe, the function is present. We know її behind formula (53) by setting x0 = y0 = z0 = 0. Then

In this rank, the function and is determined with accuracy up to a sufficient posting dodanku. Assume Z = 0, then u = xyz. Otzhe,

Let's look at the curvilinear integral

taking after a deak with a flat curve L, which connects the points Mі N. Let's assume that the functions P(x, y)і Q(x, y) may be uninterrupted private holidays in the area that you can see D. Of course, for such minds of writing a curvilinear integral cannot lie in the form of a curve L, and to deposit only in the position of the cob and end point Mі N.

Let's look at two more curves MPNі MQN, which lies near the open area D and connecting points Mі N. Come on

(1)

Then on the basis of powers 1 and 4 curvilinear integrals can be:

tobto. closed-loop integral L

In the rest of the formula, the curvilinear integral of takings along a closed contour L folded from curves MPNі NQM. Tsey contour L you can, obviously, vvazhati dovilnym.

In this rank, mind you:

so for any two points M and N the curvilinear integral does not lie in the form of a curve, but only in the form of a curve, but only lie in the position of these points, next, what curvilinear integral behind any closed contour is equal to zero .

Fair and wicked visnovok:

if the curvilinear integral behind any closed loop is equal to zero, then this curvilinear integral cannot lie in the form of a curve, which is between two points, and lie down only in the camps of tsikh dot . It's true, scho of equivalence (2) exuberant equivalence (1)

Theorem

Let the functions P(x, y), Q(x, y) be used at all points of the deco region D at once with their own private and without interruption. Then, in order to have a curvilinear integral behind any closed loop L, which lies at the edge of the room, reaching zero, then. shob

(2΄)

necessary and sufficient vikonannya equivalence

at all points of area D.

Bringing

Let's take a closer look at the closed circuit L in the region D and for the new one, we write Green's formula:

If the mind wins (3), then the underlying integral, which costs evil, is also equal to zero i, then,

in such a manner, sufficiency wash (3) brought.

Let's bring it now necessity mind, tobto. it is possible that evenness (2) is victorious for any closed curve L in the region D, then at the skin point of the region, the mind is victorious (3).

It is acceptable, on the other hand, that jealousy (2) wins, tobto.

and Umov (3) does not win, tobto.

hot bi in one point. Come on, for example, the singing point may be nervous

Since the left part of the nervousness has an uninterrupted function, it will be positive and more than a certain number at all points to reach a small area to avenge the point. Vіzmemo podvіyny іntegrа іn іy galluzі vіd raznitsi. Vin matime has a positive meaning. True,

Ale behind the Green formula, the left part of the rest of the unevenness is closer to the curvilinear integral over the inter-region, which, behind the allowances, is closer to zero. Otzhe, ostannya nerіvnіst supercheat minds (2), otzhe, pripuschennya, scho on vіdminu vіd zero wanting to be at one point, not so. Sounds screaming, what

at all points D.

Otzhe, the theorem is finished again.

At the hour of the wedding of the differential equals, it was brought to mind

is equivalent to the fact that viraz pdx + Qdyє the latest differential of the current function u(x, y), then.

Ale in tsomu vipadku vector

є gradient function u(x, y);

Function u(x, y), the gradient is similar to the vector potential which vector.

Let us know that who has a curvilinear integral Behind any curve L, which connects the points M and N, the difference between the values ​​of the function i at these points:

Bringing

Yakscho Рdx + Qdyє the top differential of the function u(x, y), then I will look at the curvilinear integral

For the calculation of this integral, we write the parametric alignment of the curve L, which connects the points Mі N:

Viraz, what to stand at the temples, function in t, which is a completely similar function according to t. Tom

Yak mi bachimo, the curvilinear integral in the form of a continuous differential cannot lie in the form of a curve, for which integration is necessary.

In this manner:

mind the independence of curvilinear integrals of the second kind form the way of integration like this:

Yakshcho in deakіy galuzі P(x, y)і Q(x, y) without interruption together with their own i, then:

1. in the area D do not lie in the form ways of integration, yakscho yogo meaning behind a shmatkovo-smooth curve, scho to lie at tsіy galuzі and mum zagalny cob і zagalny kіnets however.

2. integral uzdovzh be-like a closed curve L, which lies in the region D is equal to zero.

3. Main function u(x, y), for yakoї viraz pdx+qdyІsnuє povny differential, then.

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

5

at the skin point of the area D.

For the calculation of the integral, so as not to fall into the contour of the integration

next choose as the most navigating path of integration of the laman, that the connecting points and lankas are parallel to the axes Ox and Oy.

Pidintegral Viraz P(x, y)dx + Q(x, y)dy for the appointment of minds top differential acting functions u = u(x, y) tobto.

du(x, y) = P(x, y)dx + Q(x, y)dy

function u(x, y)(original) you can know how to calculate the most common curvilinear integral by lamania de - be it a fixed point, B(x, y) – changing point, and the point is the maximum coordinate X that . Todi vzdovzh maєmo that dy = 0, and vzdovzh maєmo x = constі dx = 0.

Let's take this formula:

Similarly, integrating lamanoy de otrimaemo

Apply

1. Calculate

This integral should be deposited along the integration contour, because

We choose as a way to integrate the laman, the lines are parallel to the coordinate axes. On the first branch:

At another location:

Otzhe,

2. Know first u, like

Come on and contour Beforeє lamana OMN. Todi

3. Know, yakscho

Here, it is impossible to take the cob point of coordinates, because at this point of function P(x, y)і Q(x, y) not assigned, for that, we take it for a point, for example,. Todi

4. Know the area, surrounded by elіps

The area of ​​the figure, ruffled in the HOW area and surrounded by a closed line C, is calculated according to the formula

,

de contour Z is bypassed in a positive direction.

Let's turn the curvilinear integra into a song, creating a change

Parameter t pass values ​​from 0 to 2?

Such a rank

3. Highest curvilinear integral over the length of the arc L yakscho L– cauliflower cycloid

TASKS ON THE THEME “CURVILINE INTEGRAL”

Option 1

De L is a triangle of the straight line of the point A(0;-2) and B(4;0) lying on the XOY plane.

vzdovzh lamanoї L:OAB, de O(0,0), A(2,0), B(4,5). Bypass the contour of the anti-year arrow.

Behind the coordinates, as L is the arc of the ellipse, which lies at the first quarter.

De L is the contour of a tricot with vertices A(1,1), B(2,2), C(1,3). Bypass the contour of the anti-year arrow.

and know yoga.

7. The force field is created by the force F(x, y), which allows more points to be fixed in the cob of coordinates and direct to the cob of coordinates. To know to the robot the strength of the field, applied to the displacement of the material point of a single mass along the arc of the parabola y2 = 8x from the point (2; 4) to the point (4; 4).

Option 2

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L is a contraction of a straight point that connects O (0; 0) and A (1; 2).

2. Calculate the curvilinear integral where L is a parabolic arc from point A(-1;1) to point B(1,1). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral yakscho L - arc stake what lies in 1 and 2 squares. Bypass the contour behind the year's arrow.

4. Zastosovuyuchi Green's formula, calculate the integral, de L - the contour, the solutions of the line and the opposite axis OX when bypassing the contour of the anti-Godinnikov arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. At the skin point of the force field, the force can be directly negative and equal to the square of the abscissa of the program point. To know the field for the robot when moving a single parabolic mass from the point (1,0) to the point (0,1).

Option 3

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

1. de L - the arc of the parabola is seen by the parabola.

2. Calculate the curvilinear integral yakscho L-wire is a straight line, connecting points A(0,1), B(2,3). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral as L is the arc of the first arc of the cycloid. Bypass the contour behind the year's arrow.

4. Zastosovuyuchi Green's formula, calculate the integral de L – elіps Obkhіd contour of the anti-godinnikov's arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. Calculate the robot's strength and hour of movement of the material point of the upper half of the ellipse from point A (a, 0), point B (-a, 0).

Option 4.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

1. de L - outline of a square

2. Calculate the curvilinear integral where L is the arc of the parabola of the point А(0,0), up to the point (1,1). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral yakscho L - the upper half of the ellipse Bypass the contour behind the year's arrow.

4. Using the Green formula, calculate the integral de L - the contour of the tricot with vertices A (1; 0), B (1; 1), C (0.1). Bypass the contour of the anti-year arrow.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. A force is applied at the skin point of the stake, with projections on the coordinate axis є Assign the force to the robot for the hour of moving the material point along the stake. Why does the robot cost zero?

Option 5.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - a straight line that connects the points 0 (0.0), і A (4; 2)

2. Calculate the curvilinear integral as L is the arc of the curved point that goes from A(0.1) to point B (-1,e). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral as L - 1 quarter of the stake Bypass the contour behind the year's arrow.

4. Zastosovuyuchi Green's formula, calculate the integral de L - contour, surroundings and bypassing the contour of the opposite arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. The field is created by force // = straight way to set the cut from the straight line radius - the vector of the point її zastosuvannya. To know the field for the robot when the material point of mass m is moved behind the arc of the stake from the point (a, 0) to the point (0, a).

Option 6

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - a quarter of a stake, which lies in the I quadrant.

2. Calculate the curvilinear integral yakcho L - laman ABC, A (1; 2), B (1; 5), C (3; 5). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral as L is the upper half of the stake Bypass the contour behind the year's arrow.

4. Zastosovuyuchi Green's formula, calculate the integral de L - the contour, the surroundings, bypassing the contour of the anti-godinnikov's arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. Know the work of the spring force, directly on the cob of coordinates, as the point of stagnation of the force describes the opposite arrow of the quarter of the ellipse what lies in the I quadrant. The magnitude of the force is proportional to the distance of the point to the cob of coordinates.

Option 7.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - part of the parabola from the point (1, 1/4) to the point (2; 1).

2. Calculate the curvilinear integral de L - vertex of the straight line, which connects the points B (1; 2) and B (2; 4). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral as L - the first arch of the cycloid By the contour behind the year hand.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. The material point of a single mass moves along the stake under the direction of force, projections of which are on the coordinate axis є . Induce strength on the cob of the skin stake. Know the work of the contour.

Option 8.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - contour of a rectangle with vertices at points 0 0 (0; 0), A (4; 0), B (4; 2), C (0; 2).

2. Calculate the curvilinear integral, for example L is the arc of the parabola from point A (0; 0) to point B (1; 2). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral yakscho L - part of a stake 1. Bypass the contour behind the annual arrow.

4. Zastosovuyuchi Green's formula, calculate the integral de L - the contour of the tricot with vertices A (0; 0), B (1; 0), C (0; 1). Bypass the contour of the anti-year arrow.

5. Install, chi vykonuetsya mind independence of the integral in the way of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. The material point moves with an ellipse pіd dієyu force, the value of which is the most expensive point to the center of the ellipse and is straightened to the center of the ellipse. Calculate the strength of the robot, as a point to bypass the entire elіps.

Option 9.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - the arc of the parabola, which lies between the points

A , (2;2).

2. Calculate the curvilinear integral where L is a constriction of a straight line, which connects the points A (5; 0) and B (0.5). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral, such as L - the arc of the ellipse between the points, which will show the circumference of the contour behind the year arrow.

4. Zastosovuyuchi Green's formula, calculate the integral de L - around the contour of the counter arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. At the skin point of the curve, a force is applied, which projections on the coordinate axes indicate the work of the force when the material point of a single mass is moved along the curve from the point M (-4; 0) to the point N (0; 2).

Option 10.

1. Calculate the curvilinear integral over the edge of the arc (Cartesian coordinates).

De L - a straight line that joins points A

2. Calculate the curvilinear integral, for example L is the arc of the curve from the point A(1;0) to B(e,5). Bypass the contour of the anti-year arrow.

3. Calculate the curvilinear integral as L is the arc of the stake what lies at 1U square. Bypass the contour behind the year's arrow.

4. Using the Green formula, calculate the integral de L - the contour of the tricot with vertices A (1; 0), B (2; 0), C (1; 2). Bypass the contour of the anti-year arrow.

5. Establish how the mind independence of the integral is calculated in the direction of integration for the integral and know yoga.

6. Revisit, chi є tasks with a new differential of the function U(x, y), and know її.

7. A force is applied to the skin point of the line, the projections of which are on the coordinate axis Calculate the robot affected by the force when the material point is moved along the line from M (1; 0) to the point N (0; 3).

Lecture 4

Topic: Green's formula. Cleanse the independence of the curvilinear integral in the way of integration.

Green's formula.

Green's formula establishes a connection between a curvilinear integral over a closed contour Г on a plane and a lower integral over an area surrounded by a contour.

The curvilinear integral over a closed contour Г is indicated by the symbol Closed contour Г begins at the main point of the contour and ends at point B. The integral along a closed contour does not lie if the point B is selected.

Appointment 1. The bypass of the contour G is considered positive, as when bypassing the contour G, the area D becomes left-handed. P + - circuit P bypasses the positive direction, P - - circuit bypasses the negative direction. at the opposite direction

G+

X

Y

c

d

X = x 1 (y)

X = x 2 (y)

a

b

B

C

Y=y 2 (x)

Y= y 1 (x)

m

n

Let's look at the underlying integral

.

Similarly, it can be argued that:

From equalities (1) and (2) it is necessary:

Otzhe,

Green's formula for crushed omissions has been completed.

Respect 1. Green's formula remains fair, as between the G region D and the real straight lines, parallel to the axis 0X or 0Y shifts more lower at two points. Krim ts'ogo, Green's formula is valid for n-stellar regions.

Wash the independence of the curvilinear integral as an integration on the plane.

In this paragraph, it’s easy to understand, in viconannists, for example, the curvilinear integral is to fall in the direction of integration, and to fall in the form of the cob and end points of integration.

Theorem 1. In order to have a curvilinear integral without lying in the path of integration in a single-link region, it is necessary and sufficient, so that the integral, of taking along a closed, shmatkovo-smooth contour in this region, reaches zero.

Proof: Necessity. It is given: to deposit in the direction of integration. It is necessary to bring that the curvilinear integral behind a closed, smooth-smooth contour is equal to zero.

Let's take some piecewise-smooth closed contour G near the area D. On the contour G, take some more points B and C.

G

D

n

m

B

C

Oskіlki lie in the path of integration, then

, then.

Prosperity. Given: Curvilinear integral Be-yakim zaknenim shmatkovo-smooth contour to zero.

It is necessary to prove that the integral should be deposited in the direction of integration.

Let's look at the curvilinear integral behind two scrambled-smooth contours that connect points B and C. Behind the mind:

Tobto. curvilinear

integral to deposit in the direction of integration.

Theorem 2. Go without interruption at the same time with private walks and in a one-link space D. In order to have a curvilinear integral not laying low in the path of integration is necessary and sufficient, so that the division D was victorious

Proof: Sufficiency. Given: . It is necessary to bring what to deposit in the direction of integration. For whom it is sufficient to bring what dovnyuє zero behind a closed, shmatkovo-smooth contour. According to Green's formula, we can:

Necessity. Given: By Theorem 1, the curvilinear integral to deposit in the direction of integration. It is necessary to bring what

6. The formula for the average value of the sing integral.

7. Integral over the changing upper boundary. Yogo bezperervnіst that differentiation.

8. Newton-Leibniz formula for the sing integral.

9. Calculation of the sing integral by parts and replacement of the change.

10. Stitching the sing integral (the area of ​​a flat figure, the length of a curved arc, the volume of the body wrapping).

11. Understanding the number series ta yogo sumi. Criteria Koshі zbіzhnostі row. Necessary intelligence.

12. Signs of Delambert and Koshі zbіzhnostі ryadіv іz nevid'єmnimi members.

13. Integral sign of Koshі zbіzhnostі number series.

14. Significant number rows. Absolutely that mental zbіzhnist. Rows of signs. Leibniz sign.

15. Functional series. The amount is low. The value of equal income is low. Criterion Koshі equal profitability of the functional series.

16. Weijerstras sign of even living.

18. Step row. Abel's theorem.

19. Radius of life of a static row. The Cauchy-Hadamard formula for the radius of the radius of the static series.

21. Functions of rich change. Understanding the n-world Euclidean space. An impersonal point of the Euclidean space. The sequence of points and її boundary. Designated functions of a small number of changes.

22. Between the functions of a number of changes. Uninterrupted function. Private holidays

23. Designation of the differential function of a number of variables and її differential. Pokhіdnі and differentials of higher orders.

24. Taylor's formula for riches of change. The extremum of the function of a small number of variables. Necessary mind extremum. Enough mind extremum.

25. Sustained integral and yoga of power. Zvedennya podvіynogo іntegral up to repeated.

27. Replacement of changes in the third integral. Cylindrical and spherical coordinates.

28. Calculation of the area of ​​a smooth surface, given parametrically and explicitly.

29. Appointment of curvilinear integrals of the first and other kind, their main power and calculation.

30. Green's formula. Cleanse the independence of the curvilinear integral in the way of integration.

31. Superficial integrals of the first and other kind, their main power and calculation.

32. Theorem of Gaus-Ostrogradsky, її notation in coordinate and vector (invariant) forms.

33. Stokes formula, written in coordinate and vector (invariant) forms.

34. Scalar and vector fields. Gradient, divergence, rotor. Potentiyne and saline fields.

35. Hamilton operator. (Nabla) yogo zastosuvannya (apply).

36. The main concepts that are used are the first-order differential equations (ode) of the first order: global and private solution, global integral, integral curve. Zavdannya Koshі, її geometrically significant.

37. Integration of an ode to the first order with the changes that are divided, and the same.

38. Integration of linear odes of the first order and equalization of Bernoulli.

39. Integration of an ode to the first fret in polar differentials. Integrating multiplier.

40. Differential equalities of the first order, invariably similar. Parameter request method.

41. Equivalence of the n-th order with constant coefficients. Characteristically equal. Fundamental solution system (FSR) of homogeneous alignment, global solution of heterogeneous alignment.

42. System of linear differential equalities of the first order. FSR of a homogeneous system. A glaring vision of a homogeneous system.

30. Green's formula. Cleanse the independence of the curvilinear integral in the way of integration.

Green's formula: If C is closed between the area D and the functions P(x, y) and Q(x, y) together with their private first-order analogs without interruption in the closed area D (including the cordon C), then Green's formula is valid:, moreover bypass around contour C is selected so that area D is left-handed.

Three lectures: Give the given functions P(x,y) and Q(x,y) as non-interrupted domains D at the same time from the private ones of the first order. Integral over the cordon (L), which is exactly to lie in the region D and cover all points in the region D: . Positive directly to the contour is such, if the part of the contour is surrounded by a left hand.

Umov’s independence of the curvilinear integral of the 2nd kind of integration path. It is necessary that sufficient mind that the curvilinear integral of the first kind, which connects the points M1 and M2, does not lie in the path of integration, but lie only in the cob and end points, equanimity:.

.

31. Superficial integrals of the first and other kind, their main power and calculation.

- Surface manager.

We project S on a plane xy, we take a board D. We draw a board D with a grid line on the part, which are called Di. From the skin point of the skin line, we draw parallel lines z, then i S is divided into Si. We add the integral sum: . We direct the maximum of the diameter Di to zero:, we take:

Ce surface integral of the first kind

This is how the surface integral of the first kind comes into play.

Appointment briefly. As a rule, there is a limiting boundary of the integral sum, so that one cannot lie in the way of splitting S on the elementary plot Si and in the choice of points, vin is called the superficial integral of the first kind.

When passing from the changed x і y to u and v:

P the surface integral can have all the power of the stellar integral. Divas are higher in food.

The purpose of the surface integral is of a different kind, which is the main power of that calculation. Link from the integral of the first kind.

Let the surface S be given, surrounded by a line L (Fig. 3.10). It is possible to add two normals to the surface S on the surface S, which can not be a double point with the cordon L. Outline the point M behind the contour L, choosing the straight line normal.

If the position of the point M rotates along this same normal (and not opposite), then the surface S is called two-sided. We can only look at two-sided surfaces. Bilateral surface - whether it be smooth surface with equal.

Let S be a two-sided open surface, surrounded by a line L, so that there is no point of self-crossing. We choose the same side of the surface. Let's call the positive direct bypass of the contour L such a straight line, with Russia on the other side of the surface, the surface itself is devoid of evil. A two-sided surface that is installed on it in such a positive order by a direct bypass of the contours is called an oriented surface.

Let's move on to a surface integral of a different kind. Let's take a double-sided surface S, which is formed from the final number of pieces, leather from some tasks equal to the mind, or a cylindrical surface with satisfying parallel axes Oz.

Let R (x, y, z) - a function, assigned and uninterrupted on the surface S. By a number of lines, S is divided by a sufficient order into n "elementary" plots ΔS1, ΔS2, ..., ΔSi, ..., ΔSn, no matter sleepy internal points. On the skin space ΔSi, we select the point Mi(xi,yi,zi) (i=1,...,n) in a reasonable order. Let (ΔSi)xy be the area of ​​the projection of the plot ΔSi onto the coordinate plane Oxy, taken with the "+" sign, so that the normal to the surface S at the point Mi(xi,yi,zi) (i=1,...,n) sets the Vіsyu Oz is a hostile cut, and with the sign "-", which means that this cut is stupid. We add the integral sum for the function R(x,y,z) over the surface S after changing x,y: . Let λ be the largest of the diameters ΔSi (i = 1, ..., n).

If there is a final boundary, so as not to lie in the way of splitting the surface S on the "elementary" plot ΔSi and choosing a point, then vin is called the surface integral along the chosen side of the surface S in the function R (x, y, z) for the coordinates x, y ( or a superficial integral of another kind) and is assigned .

Similarly, it is possible to induce surface integrals for the coordinates x, z or y, z along the opposite side of the surface, that is. і .

As well as all the integrals, you can introduce a "high" integral on the opposite side of the surface: .

A superficial integral of another kind may be dependent on the power of the integral. It is more respectful that if any surface integral of another kind changes the sign of the change of the side of the surface.

Link between surface integrals of the first and other kind.

Let the surface S be given equal: z \u003d f (x, y), moreover, f (x, y), f "x (x, y), f" y (x, y) - uninterrupted functions near the closed region τ (the projection of the surface S onto the coordinate plane Oxy), and the function R(x,y,z) is continuous on the surface S. sides of the surface S. Todi.

For a zagalny vpadku maєmo:

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