Curvilinear integral of the 1st kind over a closed contour. MA

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

We 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. Organize the division of the region-arc L into elements - elementary arcs so that these elements do not have common interior points and
(condition A )

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

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

Passing to the limit under the condition
(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 the curvilinear integral of the first kind exists as the limit of integral sums.

Comment. This limit does not depend on

method of choosing a partition, as long as condition A

selection of "marked points" on the partition elements,

method for 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) homogeneity property
.

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

2. Additivity. If a
, then
=
+

Proof. We choose a partition of the domain L so that none of the elements of the partition (initially and when the partition is refined) contains both the elements L 1 and the elements L 2 at the same time. This can be done by the existence theorem (remark on the theorem). Further, the proof is carried out in terms of integral sums, as in Section 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 pass to the limit.

Note that, in particular, it is possible

5. Estimation theorem.

If there are constants
, something

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

6. Mean 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 fulfilled. 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. Consequently,
.

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

The problem of the mass of the curve. Curvilinear integral of the 1st kind.

The problem of the mass of the curve. Let at each point of the piecewise-smooth material curve L: (AB) its density be given. Determine the mass of the curve.

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

3. Let's construct the integral sum , where is the length of the arc (usually the same designations 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 on each element and took a finite number of elements.

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

.

The existence theorem.

Let the function be continuous on a piecewise smooth arc L. Then a curvilinear 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) homogeneity property .

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

2. Additivity.
If a , then = +

3. .Here is the length of the arc .

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

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

Note that, in particular, it is possible

5. Estimation theorem.

If there are constants such that , then

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

6. Mean 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 fulfilled. 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, the function must take this value at some point. Consequently, .

Calculation of a curvilinear integral of the first kind.

We parametrize 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 curvilinear integral of the first kind reduces 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.

How much work does the force do?F(M) when moving the pointMin an arcAB? If the arc AB were a straight line segment, and the force would be constant in magnitude and direction when the point M moves along the arc AB, then the work could be calculated by the formula , where is the angle between the vectors. In the general case, this formula can be used to construct an integral sum, assuming that the force is constant on an arc element of sufficiently small length. Instead of the length of a small element of the arc, you can take the length of the chord subtending it, since these quantities are equivalent infinitesimal quantities under the condition (first semester).

1. Organize the partition of the region-arc AB into elements - elementary arcs so that these elements do not have common interior points and ( condition A )

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

3. Construct the integral sum , where is the vector directed along the chord that subtends the -arc .

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

. Often referred to

The 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

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

Selecting "marked points" on the 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) homogeneity property .

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

2. Additivity.
If a , then = + .

Proof. Let us choose a partition of the domain L so that none of the elements of the partition (initially and when the partition is refined) contains both the elements L 1 and the elements L 2 at the same time. This can be done by the existence theorem (remark on the theorem). Further, the proof is carried out in terms of integral sums, as in Section 1.

3. Orientability.

= -

Proof. The arc integral –L, i.e. in the negative direction of bypassing the arc, there is a limit of integral sums, in the terms of which there is () instead. Taking out the "minus" from the scalar product and from the sum of a finite number of terms, passing to the limit, we obtain the required result.

Theoretical minimum

Curvilinear and surface integrals often occur in physics. They come in two varieties, the first of which is discussed here. This
the type of integrals is constructed according to the general scheme, according to 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 of the parts. At each of these points, the value of the integrand is calculated and multiplied by the measure of the part that
the given point belongs (the length of the segment, the area or volume of the partial area). Then all such products are summed, and the limit
transition to partitioning an object into infinitely small parts. The resulting limit is called the integral.

1. Definition of a curvilinear integral of the first kind

Consider a function defined on a curve . The curve is assumed to be rectifiable. Recall what this means, roughly speaking,
that a polyline with arbitrarily small links can be inscribed in a curve, and in the limit of an infinitely large number of links, the length of the polyline must remain
final. The curve is divided into partial arcs of length and a point is selected on each of the arcs. The work is being compiled
summation 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, which follows directly from its definition, is independence from the direction of integration, i.e.
.

2. Definition of a surface integral of the first kind

Consider a function defined on a smooth or piecewise smooth surface . The surface is broken into partial regions
with areas , a point is selected in each such area. A work is being compiled , the summation
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 the curvilinear integral of the first kind can already be seen from its formal notation, but in fact it follows directly from
definitions. The integral is reduced to a definite one, only it is necessary to write down the differential of the arc of the curve along which the integration is carried out.
Let's start with a simple case of integration along a plane curve given by an explicit equation . In this case, the arc differential
.
Then, in the integrand, the variable is changed, and the integral takes the form
,
where the segment corresponds to the change in the variable along that part of the curve over which the integration is carried out.

Very often the curve is set parametrically, i.e. type equations. Then the arc differential
.
This formula is very easy to justify. Basically, it's the Pythagorean theorem. The arc differential is actually the length of an infinitesimal part of the curve.
If the curve is smooth, then its infinitesimal part can be considered rectilinear. For a straight line, the relation
.
In order for it to be carried out for a small arc of the curve, one should pass from finite increments to differentials:
.
If the curve is given parametrically, then the differentials are simply calculated:
etc.
Accordingly, after changing the variables in the integrand, the curvilinear integral is calculated as follows:
,
where the part of the curve along which the integration is carried out corresponds to the segment of the change in the parameter .

The situation is somewhat more complicated when the curve is specified in curvilinear coordinates. This question is usually discussed in the framework of the differential
geometry. Let's give a formula for calculating the integral along the curve given in polar coordinates by the equation:
.
Let us also give a justification for the arc differential in polar coordinates. Detailed discussion of gridding the polar coordinate system
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
arcs again, you can apply the Pythagorean theorem and write:
.
From here follows the desired expression for the differential of the arc.

From a purely theoretical point of view, it is quite easy to understand that the curvilinear integral of the first kind must be reduced to its special case -
a certain integral. Indeed, making a change dictated by the parametrization 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 the reduction to the integral
along a straight line coinciding with the 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 a surface element,
over which the integration is performed. Again, let's start with the simple case of a surface given by an explicit equation . Then
.
A change is made in the integrand, and the surface integral is reduced to the double integral:
,
where is the region of the plane into which a part of the surface is projected over which the integration is carried out.

However, it is often impossible to specify a surface by an explicit equation, and then it is specified parametrically, i.e. equations of the form
.
The surface element in this case is written more complicated:
.
The surface integral is written in the corresponding way:
,
where is the range of parameters corresponding to the part of the surface over which the integration is carried out.

5. The physical meaning of the curvilinear and surface integrals of the first kind

The integrals under discussion 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 the curve is , then its mass
, where is any point of the selected piece of the curve (any, since the density is within
of this piece is assumed to be approximately constant). Accordingly, the mass of the entire curve is obtained by summing the masses of its individual parts:
.
In order for the equality to become exact, one should pass to the limit of splitting the curve into infinitely small parts, but this is the curvilinear integral of the first kind.

Similarly, the question of the total charge of the curve is resolved if the linear charge density is known .

These considerations are easily transferred to the case of an unevenly charged surface with a surface charge density . Then
the surface charge is a surface integral of the first kind
.

Remark . A cumbersome formula for a surface element given parametrically is inconvenient for memorization. Another expression is obtained in differential geometry,
it uses the so-called. the first quadratic form of the surface.

Examples of calculating curvilinear integrals of the first kind

Example 1 Integral along a line.
Calculate Integral

along the 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 the plane.
Calculate Integral

along the arc of a parabola from point to point.

The given points and allow us to express the variable from the parabola equation: .

We calculate the integral:
.

However, it was possible to carry out calculations in another way, using the fact that the curve is given by an equation that is solved with respect to the variable .
If we take a variable as a parameter, then this will lead to a slight change in the expression for the arc differential:
.
Accordingly, the integral will change somewhat:
.
This integral is easily calculated by subsuming the variable under the differential. The result is the same integral as in the first method of calculation.

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

along the top half of the circumference .

You can, of course, express one of the variables from the circle equation, 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 changing the parameter within . Calculate the arc differential:
.
In this way,

Example 4 Integral along a curve in a plane given in polar coordinates.
Calculate Integral

along the right lobe of the lemniscate .

The drawing above shows a lemniscate. Integration should 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 hold, 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