Calculate the curvilinear integral of the 1st kind along a circle. The curve is given in Cartesian rectangular coordinates

Theoretical minimum

Curvilinear and surface integrals are often found in physics. They come in two types, the first of which is discussed here. This
the type of integrals is constructed according to general scheme, by which definite, double and triple integrals are introduced. Let us briefly recall this scheme.
There is some object over which integration is carried out (one-dimensional, two-dimensional or three-dimensional). This object is broken into small parts,
a point is selected in each part. At each of these points, the value of the integrand is calculated and multiplied by the measure of the part that
belongs given point(length of a segment, area or volume of a partial region). Then all such products are summed and the limit is satisfied
transition to breaking the object into infinitesimal parts. The resulting limit is called the integral.

1. Definition of a curvilinear integral of the first kind

Let's consider a function defined on a curve. The curve is assumed to be rectifiable. Let us recall what this means, roughly speaking,
that a broken line with arbitrarily small links can be inscribed into a curve, and in the limit it is infinite large number links, the length of the broken line should remain
final. The curve is divided into partial arcs of length and a point is selected on each of the arcs. A work is being compiled
summation is carried out over all partial arcs . Then the passage to the limit is carried out with the tendency of the length of the greatest
from partial arcs to zero. The limit is a curvilinear integral of the first kind
.
An important feature of this integral, directly following from its definition, is its independence from the direction of integration, i.e.
.

2. Definition of surface integral of the first kind

Consider a function defined on a smooth or piecewise smooth surface. The surface is divided into partial areas
with areas, a point is selected in each such area. A work is being compiled , summation is carried out
over all partial areas . Then the passage to the limit is carried out with the tendency of the diameter of the largest of all partial
areas to zero. The limit is a surface integral of the first kind
.

3. Calculation of a curvilinear integral of the first kind

The method for calculating a curvilinear integral of the first kind can be seen already from its formal notation, but in fact follows directly from
definitions. The integral is reduced to a definite one; you just need to write down the differential of the arc of the curve along which the integration is carried out.
Let's start with simple case integration along a plane curve given by an explicit equation. In this case, the arc differential
.
Then a change of variable is performed in the integrand, and the integral takes the form
,
where the segment corresponds to the change in the variable along that part of the curve along which the integration is carried out.

Very often the curve is specified parametrically, i.e. equations of the form Then the arc differential
.
This formula is very simply justified. Essentially, this is the Pythagorean theorem. The arc differential is actually the length of the infinitesimal part of the curve.
If the curve is smooth, then its infinitesimal part can be considered rectilinear. For a straight line we have the relation
.
In order for it to be carried out for a small arc of the curve, one should move from finite increments to differentials:
.
If the curve is specified parametrically, then the differentials are simply calculated:
etc.
Accordingly, after changing variables in the integrand, the line integral is calculated as follows:
,
where the part of the curve along which the integration is carried out corresponds to the segment of the parameter change.

The situation is somewhat more complicated in the case when the curve is specified in curvilinear coordinates. This issue is usually discussed within the framework of differential
geometry. Let us give a formula for calculating the integral along the curve given in polar coordinates equation:
.
Let us give a justification for the differential of the arc in polar coordinates. Detailed discussion of grid construction polar system coordinates
cm. . Let us select a small arc of the curve located in relation to the coordinate lines as shown in Fig. 1. Due to the smallness of all those featured
arc again we can apply the Pythagorean theorem and write:
.
From here follows the desired expression for the differential of the arc.

With pure theoretical point From a visual perspective, it is enough to simply understand that a curvilinear integral of the first kind must be reduced to its particular case -
to a definite integral. Indeed, by making the change dictated by the parameterization of the curve along which the integral is calculated, we establish
one-to-one mapping between a part of a given curve and a segment of parameter change. And this is a reduction to the integral
along a straight line coinciding with coordinate axis- a definite integral.

4. Calculation of the surface integral of the first kind

After the previous point, it should be clear that one of the main parts of calculating a surface integral of the first kind is writing the surface element,
over which the integration is performed. Again, let's start with the simple case of a surface defined by an explicit equation. Then
.
A substitution is made in the integrand, and the surface integral is reduced to a double:
,
where is the region of the plane into which the part of the surface over which integration is carried out is projected.

However, it is often impossible to define a surface by an explicit equation, and then it is defined parametrically, i.e. equations of the form
.
The surface element in this case is written more complicated:
.
The surface integral can be written accordingly:
,
where is the area of ​​change of parameters corresponding to the part of the surface over which integration is carried out.

5. Physical meaning of curvilinear and surface integrals of the first kind

The integrals discussed have a very simple and clear physical meaning. Let there be some curve whose linear density is not
constant, and is a function of the point . Let's find the mass of this curve. Let's break the curve into many small elements,
within which its density can be approximately considered constant. If the length of a small piece of a curve is equal to , then its mass
, where is any point of the selected piece of the curve (any, since the density is within
this piece is approximately assumed to be constant). Accordingly, the mass of the entire curve is obtained by summing the masses of its individual parts:
.
For the equality to become accurate, one must go to the limit of dividing the curve into infinitesimal parts, but this is a curvilinear integral of the first kind.

The question of the total charge of the curve is resolved similarly if the linear charge density is known .

These arguments can easily be transferred to the case of a nonuniformly charged surface with surface density charge . Then
the surface charge is a surface integral of the first kind
.

Note. A cumbersome formula for a surface element defined parametrically is inconvenient to remember. Another expression is obtained in differential geometry,
it uses the so-called first quadratic form surfaces.

Calculation examples curvilinear integrals first kind

Example 1. Integral along a line.
Calculate integral

along a line segment passing through the points and .

First, we write the equation of the straight line along which the integration is carried out: . Let's find an expression for:
.
We calculate the integral:

Example 2. Integral along a curve in a plane.
Calculate integral

along a parabola arc from point to point.

Setpoints and allow you to express a variable from the parabola equation: .

We calculate the integral:
.

However, it was possible to carry out calculations in another way, taking advantage of the fact that the curve is given by an equation resolved with respect to the variable.
If we take a variable as a parameter, this will lead to small change expressions for the arc differential:
.
Accordingly, the integral will change slightly:
.
This integral is easily calculated by substituting the variable under the differential. The result is the same integral as in the first calculation method.

Example 3. Integral along a curve in a plane (using parametrization).
Calculate integral

along the top half of the circle .

You can, of course, express one of the variables from the equation of a circle, and then carry out the rest of the calculations in the standard way. But you can also use
parametric curve specification. As you know, a circle can be defined by equations. Upper semicircle
corresponds to a change in the parameter within . Let's calculate the arc differential:
.
Thus,

Example 4. Integral along a curve on a plane specified in polar coordinates.
Calculate integral

along the right lobe of the lemniscate .

The drawing above shows a lemniscate. Integration must be carried out along its right lobe. Let's find the arc differential for the curve :
.
The next step is to determine the limits of integration over the polar angle. It is clear that the inequality must be satisfied, and therefore
.
We calculate the integral:

Example 5. Integral along a curve in space.
Calculate integral

along the turn of the helix corresponding to the limits of parameter change

For the case when the domain of integration is a segment of a certain curve lying in a plane. The general notation for a line integral is as follows:

Where f(x, y) is a function of two variables, and L- curve, along a segment AB which integration takes place. If the integrand is equal to one, then the line integral equal to length arc AB .

As always in integral calculus, a curvilinear integral is understood as the limit of the integral sums of some very small parts of something very large. What is summed up in the case of curvilinear integrals?

Let there be a segment on the plane AB some curve L, and a function of two variables f(x, y) defined at the points of the curve L. Let us perform the following algorithm with this segment of the curve.

1. Split curve AB into parts with dots (pictures below).
2. Freely select a point in each part M.
3. Find the value of the function at selected points.
4. Function values ​​multiply by
   • lengths of parts in case curvilinear integral of the first kind ;
   • projections of parts onto the coordinate axis in the case curvilinear integral of the second kind .
5. Find the sum of all products.
6. Find the limit of the found integral sum provided that the length of the longest part of the curve tends to zero.

If the mentioned limit exists, then this the limit of the integral sum and is called the curvilinear integral of the function f(x, y) along the curve AB .

first kind

Case of a curvilinear integral
second kind

Let us introduce the following notation.

Mi( ζ i ; η i)- a point with coordinates selected on each site.

fi( ζ i ; η i)- function value f(x, y) at the selected point.

Δ si- length of part of a curve segment (in the case of a curvilinear integral of the first kind).

Δ xi- projection of part of the curve segment onto the axis Ox(in the case of a curvilinear integral of the second kind).

d= maxΔ s i- the length of the longest part of the curve segment.

Curvilinear integrals of the first kind

Based on the above about the limit of integral sums, a curvilinear integral of the first kind is written as follows:

.

A line integral of the first kind has all the properties that it has definite integral. However, there is one important difference. For a definite integral, when the limits of integration are swapped, the sign changes to the opposite:

In the case of a curvilinear integral of the first kind, it does not matter which point of the curve AB (A or B) is considered the beginning of the segment, and which one is the end, that is

.

Curvilinear integrals of the second kind

Based on what has been said about the limit of integral sums, a curvilinear integral of the second kind is written as follows:

.

In the case of a curvilinear integral of the second kind, when the beginning and end of a curve segment are swapped, the sign of the integral changes:

.

When compiling the integral sum of a curvilinear integral of the second kind, the values ​​of the function fi( ζ i ; η i) can also be multiplied by the projection of parts of a curve segment onto the axis Oy. Then we get the integral

.

In practice, the union of curvilinear integrals of the second kind is usually used, that is, two functions f = P(x, y) And f = Q(x, y) and integrals

,

and the sum of these integrals

called general curvilinear integral of the second kind .

Calculation of curvilinear integrals of the first kind

The calculation of curvilinear integrals of the first kind is reduced to the calculation of definite integrals. Let's consider two cases.

Let a curve be given on the plane y = y(x) and a curve segment AB corresponds to a change in variable x from a to b. Then at the points of the curve the integrand function f(x, y) = f(x, y(x)) ("Y" must be expressed through "X"), and the differential of the arc and the line integral can be calculated using the formula

.

If the integral is easier to integrate over y, then from the equation of the curve we need to express x = x(y) (“x” through “y”), where we calculate the integral using the formula

.

Example 1.

Where AB- straight line segment between points A(1; −1) and B(2; 1) .

Solution. Let's make an equation of a straight line AB, using the formula (equation of a line passing through two given points A(x1 ; y 1 ) And B(x2 ; y 2 ) ):

From the straight line equation we express y through x :

Then and now we can calculate the integral, since we only have “X’s” left:

Let a curve be given in space

Then at the points of the curve the function must be expressed through the parameter t() and arc differential , therefore the curvilinear integral can be calculated using the formula

Similarly, if a curve is given on the plane

,

then the curvilinear integral is calculated by the formula

.

Example 2. Calculate line integral

Where L- part of a circle line

located in the first octant.

Solution. This curve is a quarter of a circle line located in the plane z= 3 . It corresponds to the parameter values. Because

then the arc differential

Let us express the integrand function through the parameter t :

Now that we have everything expressed through a parameter t, we can reduce the calculation of this curvilinear integral to a definite integral:

Calculation of curvilinear integrals of the second kind

Just as in the case of curvilinear integrals of the first kind, the calculation of integrals of the second kind is reduced to the calculation of definite integrals.

The curve is given in Cartesian rectangular coordinates

Let a curve on a plane be given by the equation of the function “Y”, expressed through “X”: y = y(x) and the arc of the curve AB corresponds to change x from a to b. Then we substitute the expression of the “y” through “x” into the integrand and determine the differential of this expression of the “y” with respect to “x”: . Now that everything is expressed in terms of “x”, the line integral of the second kind is calculated as a definite integral:

A curvilinear integral of the second kind is calculated similarly when the curve is given by the equation of the “x” function expressed through the “y”: x = x(y) , . In this case, the formula for calculating the integral is as follows:

Example 3. Calculate line integral

, If

A) L- straight segment O.A., Where ABOUT(0; 0) , A(1; −1) ;

b) L- parabola arc y = x² from ABOUT(0; 0) to A(1; −1) .

a) Let’s calculate the curvilinear integral over a straight line segment (blue in the figure). Let’s write the equation of the straight line and express “Y” through “X”:

.

We get dy = dx. We solve this curvilinear integral:

b) if L- parabola arc y = x² , we get dy = 2xdx. We calculate the integral:

In the example just solved, we got the same result in two cases. And this is not a coincidence, but the result of a pattern, since this integral satisfies the conditions of the following theorem.

Theorem. If the functions P(x,y) , Q(x,y) and their partial derivatives are continuous in the region D functions and at points in this region the partial derivatives are equal, then the curvilinear integral does not depend on the path of integration along the line L located in the area D .

The curve is given in parametric form

Let a curve be given in space

.

and into the integrands we substitute

expressing these functions through a parameter t. We get the formula for calculating the curvilinear integral:

Example 4. Calculate line integral

,

If L- part of an ellipse

meeting the condition y ≥ 0 .

Solution. This curve is the part of the ellipse located in the plane z= 2 . It corresponds to the parameter value.

we can represent the curvilinear integral in the form of a definite integral and calculate it:

If a curve integral is given and L is a closed line, then such an integral is called a closed-loop integral and is easier to calculate using Green's formula .

More examples of calculating line integrals

Example 5. Calculate line integral

Where L- a straight line segment between the points of its intersection with the coordinate axes.

Solution. Let us determine the points of intersection of the straight line with the coordinate axes. Substituting a straight line into the equation y= 0, we get ,. Substituting x= 0, we get ,. Thus, the point of intersection with the axis Ox - A(2; 0) , with axis Oy - B(0; −3) .

From the straight line equation we express y :

.

, .

Now we can represent the line integral as a definite integral and start calculating it:

In the integrand we select the factor , and move it outside the integral sign. In the resulting integrand we use subscribing to the differential sign and finally we get it.

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

Curve mass problem. Let at each point of a piecewise smooth material curve L: (AB) its density be specified. Determine the mass of the curve.

Let us proceed in the same way as we did when determining the mass of a flat region ( double integral) and a spatial body (triple integral).

1. We organize the partition of the area-arc L into elements - elementary arcs so that these elements do not have common internal points And
(condition A )

2. Let us mark the “marked points” M i on the elements of the partition and calculate the values ​​of the function in them

3. Let's construct the integral sum
, Where - arc length (usually the same notations are introduced for the arc and its length). This is an approximate value for the mass of the curve. The simplification is that we assumed the arc density to be constant at each element and took a finite number of elements.

Moving to the limit provided
(condition B ), we obtain a curvilinear integral of the first kind as the limit of integral sums:

.

Existence theorem 10 .

Let the function
is continuous on a piecewise smooth arc L 11. Then a line integral of the first kind exists as the limit of integral sums.

Comment. This limit does not depend on

method for choosing a partition, as long as condition A is satisfied

selecting “marked points” on partition elements,

method of refining the partition, as long as condition B is satisfied

Properties of a curvilinear integral of the first kind.

1. Linearity a) superposition property

b) property of homogeneity
.

Proof. Let us write down the integral sums for the integrals on the left sides of the equalities. Since the integral sum has a finite number of terms, we move on to integral sums for the right-hand sides of the equalities. Then we pass to the limit, using the theorem on passage to the limit in equality, we obtain the desired result.

2. Additivity. If
, That
=
+

Proof. Let us choose a partition of the region L so that none of the elements of the partition (initially and when refining the partition) contains both elements L 1 and elements L 2 at the same time. This can be done using the existence theorem (remark to the theorem). Next, the proof is carried out through integral sums, as in paragraph 1.

3.
.Here – arc length .

4. If on an arc the inequality is satisfied, then

Proof. Let us write down the inequality for the integral sums and move on to the limit.

Note that, in particular, it is possible

5. Estimation theorem.

If constants exist
, something

Proof. Integrating inequality
(property 4), we get
. By property 1 of the constant
can be taken out from under the integrals. Using property 3, we obtain the desired result.

6. Mean value theorem(the value of the integral).

There is a point
, What

Proof. Since the function
continuous on a closed limited set, then it exists bottom edge
and top edge
. The inequality is satisfied. Dividing both sides by L, we get
. But the number
enclosed between the lower and upper bounds of the function. Since the function
is continuous on a closed bounded set L, then at some point
the function must accept this value. Hence,
.

Lecture 5 Curvilinear integrals of the 1st and 2nd kind, their properties..

Curve mass problem. Curvilinear integral of the 1st kind.

Curve mass problem. Let at each point of a piecewise smooth material curve L: (AB) its density be specified. Determine the mass of the curve.

Let us proceed in the same way as we did when determining the mass of a flat region (double integral) and a spatial body ( triple integral).

1. We organize the partition of the arc region L into elements - elementary arcs so that these elements do not have common internal points and( condition A )

3. Construct the integral sum , where is the length of the arc (usually the same notation is introduced for the arc and its length). This - approximate value mass curve. The simplification is that we assumed the arc density to be constant at each element and took a finite number of elements.

Moving to the limit provided (condition B ), we obtain a curvilinear integral of the first kind as the limit of integral sums:

.

Existence theorem.

Let the function be continuous on a piecewise smooth arc L. Then a line integral of the first kind exists as the limit of integral sums.

Comment. This limit does not depend on

Properties of a curvilinear integral of the first kind.

1. Linearity
a) superposition property

b) property of homogeneity .

Proof. Let us write down the integral sums for the integrals on the left sides of the equalities. Since the integral sum has a finite number of terms, we move on to integral sums for the right-hand sides of the equalities. Then we pass to the limit, using the theorem on passage to the limit in equality, we obtain the desired result.

2. Additivity.
If , That = +

3. Here is the arc length.

4. If the inequality is satisfied on the arc, then

Proof. Let us write down the inequality for the integral sums and move on to the limit.

Note that, in particular, it is possible

5. Estimation theorem.

If there are constants that, then

Proof. Integrating inequality (property 4), we get . By property 1, constants can be removed from integrals. Using property 3, we obtain the desired result.

6. Mean value theorem(the value of the integral).

There is a point , What

Proof. Since the function is continuous on a closed bounded set, then its infimum exists and top edge . The inequality is satisfied. Dividing both sides by L, we get . But the number enclosed between the lower and upper bounds of the function. Since the function is continuous on a closed bounded set L, then at some point the function must take this value. Hence, .

Calculation of a curvilinear integral of the first kind.

Let us parameterize the arc L: AB x = x(t), y = y(t), z =z (t). Let t 0 correspond to point A, and t 1 correspond to point B. Then the line integral of the first kind is reduced to a definite integral ( - the formula known from the 1st semester for calculating the differential of the arc length):

Example. Calculate the mass of one turn of a homogeneous (density equal to k) helix: .

Curvilinear integral of the 2nd kind.

The problem of the work of force.

1. We organize the division of the region-arc AB into elements - elementary arcs so that these elements do not have common internal points and( condition A )

2. Let us mark the “marked points” M i on the elements of the partition and calculate the values ​​of the function in them

3. Let's construct the integral sum , where is the vector directed along the chord subtending the -arc .

4. Going to the limit provided (condition B ), we obtain a curvilinear integral of the second kind as the limit of integral sums (and the work of force):

. Often denoted

Existence theorem.

Let the vector function be continuous on a piecewise smooth arc L. Then a curvilinear integral of the second kind exists as the limit of integral sums.

.

Comment. This limit does not depend on

Method for choosing a partition, as long as condition A is satisfied

Selecting “marked points” on partition elements,

A method for refining the partition, as long as condition B is satisfied

Properties of a curvilinear integral of the 2nd kind.

1. Linearity
a) superposition property

b) property of homogeneity .

Proof. Let us write down the integral sums for the integrals on the left sides of the equalities. Since the number of terms in an integral sum is finite, using the property of the scalar product, we move on to integral sums for the right-hand sides of the equalities. Then we pass to the limit, using the theorem on passage to the limit in equality, we obtain the desired result.

2. Additivity.
If , That = + .

Proof. Let us choose a partition of the region L so that none of the partition elements (initially and when refining the partition) contains both elements L 1 and elements L 2 at the same time. This can be done using the existence theorem (remark to the theorem). Next, the proof is carried out through integral sums, as in paragraph 1.

3. Orientability.

= -

Proof. Integral over the arc –L, i.e. V negative direction traversal of the arc is the limit of integral sums in the terms of which there is (). Taking the “minus” out of the scalar product and from the sum finite number terms, passing to the limit, we obtain the required result.