1st kind.

1.1.1. Definition of a curvilinear integral of the 1st kind

Let on the plane Oxy given curve (L). Let for any point of the curve (L) continuous function defined f(x;y). Let's break the arc AB lines (L) dots A=P 0, P 1, P n = B on n arbitrary arcs P i -1 P i with lengths ( i = 1, 2, n) (Fig. 27)

Let's choose on each arc P i -1 P i arbitrary point M i (x i ; y i) , let's calculate the value of the function f(x;y) at the point M i. Let's make an integral sum

Let where.

λ→0 (n→∞), independent of the method of partitioning the curve ( L)to elementary parts, nor from the choice of points M i curvilinear integral of the 1st kind from function f(x;y)(curvilinear integral along the length of the arc) and denote:

Comment. The definition of the curvilinear integral of the function is introduced in a similar way f(x;y;z) along the spatial curve (L).

Physical meaning of a curvilinear integral of the 1st kind:

If (L)- flat curve with a linear plane, then the mass of the curve is found by the formula:

1.1.2. Basic properties of a curvilinear integral of the 1st kind:

3. If the integration path is divided into parts such that , and have a single common point, then .

4. Curvilinear integral of the 1st kind does not depend on the direction of integration:

5. , where is the length of the curve.

1.1.3. Calculation of a curvilinear integral of the 1st kind.

The calculation of a curvilinear integral is reduced to the calculation of a definite integral.

1. Let the curve (L) is given by the equation. Then

That is, the arc differential is calculated using the formula.

Example

Calculate the mass of a straight line segment from a point A(1;1) to the point B(2;4), If .

Solution

Equation of a line passing through two points: .

Then the equation of the line ( AB): , .

Let's find the derivative.

Then . = .

2. Let the curve (L) specified parametrically: .

Then, that is, the arc differential is calculated using the formula.

For the spatial case of specifying a curve: Then

That is, the arc differential is calculated using the formula.

Example

Find the arc length of the curve, .

Solution

We find the length of the arc using the formula: .

To do this, we find the arc differential.

Let's find the derivatives , , . Then the length of the arc: .

3. Let the curve (L) specified in the polar coordinate system: . Then

That is, the arc differential will be calculated using the formula.

Example

Calculate the mass of the line arc, 0≤ ≤ if .

Solution

We find the mass of the arc using the formula:

To do this, let's find the arc differential.

Let's find the derivative.

1.2. Curvilinear integral of the 2nd kind

1.2.1. Definition of a curvilinear integral of the 2nd kind


Let on the plane Oxy given curve (L). Let on (L) a continuous function is given f(x;y). Let's break the arc AB lines (L) dots A = P 0 , P 1 , P n = B in the direction from the point A to the point IN on n arbitrary arcs P i -1 P i with lengths ( i = 1, 2, n) (Fig. 28).

Let's choose on each arc P i -1 P i arbitrary point M i (x i ; y i), let's calculate the value of the function f(x;y) at the point M i. Let's make an integral sum, where - arc projection length P i -1 P i per axis Oh. If the direction of movement along the projection coincides with the positive direction of the axis Oh, then the projection of the arcs is considered positive, otherwise - negative.

Let where.

If there is a limit on the integral sum at λ→0 (n→∞), independent of the method of partitioning the curve (L) into elementary parts, nor from the choice of points M i in each elementary part, then this limit is called curvilinear integral of the 2nd kind from function f(x;y)(curvilinear integral over the coordinate X) and denote:

Comment. The curvilinear integral over the y coordinate is introduced similarly:

Comment. If (L) is a closed curve, then the integral over it is denoted

Comment. If on ( L) three functions are given at once and from these functions there are integrals , , ,

then the expression: + + is called general curvilinear integral of the 2nd kind and write down:

1.2.2. Basic properties of a curvilinear integral of the 2nd kind:

3. When the direction of integration changes, the curvilinear integral of the 2nd kind changes its sign.

4. If the integration path is divided into parts such that , and have a single common point, then

5. If the curve ( L) lies in the plane:

Perpendicular axis Oh, then =0;

Perpendicular axis Oy, That ;

Perpendicular axis Oz, then =0.

6. A curvilinear integral of the 2nd kind over a closed curve does not depend on the choice of the starting point (depends only on the direction of traversing the curve).

1.2.3. Physical meaning of a curvilinear integral of the 2nd kind.

Job A forces when moving a material point of unit mass from a point M exactly N along ( MN) is equal to:

1.2.4. Calculation of a curvilinear integral of the 2nd kind.

The calculation of a curvilinear integral of the 2nd kind is reduced to the calculation of a definite integral.

1. Let the curve ( L) is given by the equation .

Example

Calculate where ( L) - broken line OAB: O(0;0), A(0;2), B(2;4).

Solution

Since (Fig. 29), then

1)Equation (OA): , ,

2) Equation of a line (AB): .

2. Let the curve (L) specified parametrically: .

Comment. In the spatial case:

Example

Calculate

Where ( AB)- segment from A(0;0;1) before B(2;-2;3).

Solution

Let's find the equation of the line ( AB):

Let's move on to the parametric recording of the equation of a straight line (AB). Then .

Point A(0;0;1) corresponds to the parameter t equal: therefore t=0.

Point B(2;-2;3) corresponds to the parameter t, equal: therefore, t=1.

When moving from A To IN,parameter t changes from 0 to 1.

1.3. Green's formula. L) incl. M(x;y;z) with axles Ox, Oy, Oz

Calculation of a curvilinear integral over coordinates.

The calculation of a curvilinear integral over coordinates is reduced to the calculation of an ordinary definite integral.

Consider the curvilinear integral of the 2nd kind under the arc:

(1)

Let the equation of the integration curve be given in parametric form:

Where t- parameter.

Then from equations (2) we have:

From the same equations written for points A And IN,

let's find the values t A And t B parameters corresponding to the beginning and end of the integration curve.

Substituting expressions (2) and (3) into integral (1), we obtain a formula for calculating a curvilinear integral of the 2nd kind:

If the integration curve is given explicitly with respect to the variable y, i.e. as

y=f(x), (6)

then we accept the variable x per parameter (t=x) and we obtain the following entry of equation (6) in parametric form:

From here we have: , t A =x A , t B =x B, and the curvilinear integral of the 2nd is reduced to a definite integral over the variable x:

Where y(x)– equation of the line along which integration is performed.

If the equation of the integration curve AB specified explicitly relative to the variable x, i.e. as

x=φ(y) (8)

then we take the variable as a parameter y, we write equation (8) in parametric form:

We get: , t A =y A , t B =y B, and the formula for calculating the integral of the 2nd kind will take the form:

Where x(y)– line equation AB.

Notes.

1). A curvilinear integral over coordinates exists, i.e. there is a finite limit on the integral sum at n→∞ , if on the integration curve of the function P(x, y) And Q(x,y) are continuous, and the functions x(t) And y(t) are continuous along with their first derivatives and .

2). If the integration curve is closed, then you need to follow the direction of integration, since

Calculate integral , If AB given by the equations:

A). (x-1) 2 +y 2 =1.

b). y=x

V). y=x 2

Case A. The line of integration is a circle of radius R=1 centered at a point C(1;0). Its parametric equation is:

We find

Let's determine the parameter values t at points A And IN.

Point A. t A =π .

Case B. The line of integration is a parabola. We accept x per parameter. Then , , .

We get:

Green's formula.

Green's formula establishes a connection between a curvilinear integral of the 2nd kind over a closed contour and a double integral over a region D, limited by this contour.

If the function P(x, y) And Q(x, y) and their partial derivatives are continuous in the region D, limited by contour L, then the formula holds:

(1)

    - Green's formula.

Proof.

Consider in the plane xOy region D, correct in the direction of the coordinate axes Ox And Oy.

TO ontur L straight x=a And x=b is divided into two parts, on each of which y is a single-valued function of x. Let the upper section ADV contour is described by the equation y=y 2 (x), and the lower section DIA contour - equation y=y 1 (x).

Consider the double integral

Considering that the inner integral is calculated at x=const we get:

.

But the first integral in this sum, as follows from formula (7), is a curvilinear integral along the line ACA, because y=y 2 (x)– the equation of this line, i.e.

and the second integral is the curvilinear integral of the function P(x, y) along the line DIA, because y=y 1 (x)– equation of this line:

.

The sum of these integrals is a curvilinear integral over a closed loop L from function P(x, y) by coordinate x.

As a result we get:

(2)

Breaking the outline L straight y=c And y=d to plots GARDEN And SVD, described respectively by the equations x=x 1 (y) And x=x 2 (y) similarly we get:

Adding the right and left sides of equalities (2) and (3), we obtain Green’s formula:

.

Consequence.

Using a curvilinear integral of the 2nd kind, you can calculate the areas of plane figures.

Let's determine what the functions should be for this P(x, y) And Q(x, y). Let's write down:

or, using Green's formula,

Therefore, the equality must be satisfied

what is possible, for example, with

Where do we get:

(4)

Calculate the area enclosed by an ellipse whose equation is given in parametric form:

Condition for the independence of the curvilinear integral over coordinates from the integration path.

We have established that, in a mechanical sense, a curvilinear integral of the 2nd kind represents the work of a variable force on a curvilinear path, or in other words, the work of moving a material point in a field of forces. But it is known from physics that work in the field of gravity does not depend on the shape of the path, but depends on the position of the starting and ending points of the path. Consequently, there are cases when a curvilinear integral of the 2nd kind does not depend on the path of integration.

Let us determine the conditions under which the curvilinear integral over coordinates does not depend on the path of integration.

Let in some area D functions P(x, y) And Q(x, y) and partial derivatives

And continuous. Let us take points in this area A And IN and connect them with arbitrary lines DIA And AFB.

If a curvilinear integral of the 2nd kind does not depend on the path of integration, then

,

(1)

But integral (1) is a closed loop integral ACBFA.

Consequently, a curvilinear integral of the 2nd kind in some region D does not depend on the path of integration if the integral over any closed contour in this region is equal to zero.

Let us determine what conditions the function must satisfy P(x, y) And Q(x, y) in order for equality to be satisfied

, (2)

those. so that the curvilinear integral over coordinates does not depend on the path of integration.

Let in the area D functions P(x, y) And Q(x, y) and their partial derivatives are first order and continuous. Then, in order for the curvilinear integral over the coordinates

does not depend on the path of integration, it is necessary and sufficient that at all points of the region D equality was satisfied

Proof.

Consequently, equality (2) is satisfied, i.e.

, (5)

for which it is necessary to fulfill condition (4).

Then from equation (5) it follows that equality (2) is satisfied and, therefore, the integral does not depend on the path of integration.

Thus, the theorem is proven.

Let us show that the condition

is satisfied if the integrand

is the complete differential of some function U(x, y).

The total differential of this function is equal to

. (7)

Let the integrand (6) be the total differential of the function U(x, y), i.e.

whence it follows that

From these equalities we find expressions for the partial derivatives and:

, .

But the second mixed partial derivatives do not depend on the order of differentiation, therefore, which was what needed to be proved. curvilinear integrals. It should also... applications. From theory curvilinear integrals it is known that curvilinear integral of the form (29 ...

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    • Definition: Let at each point of a smooth curve L=AB in the plane Oxy a continuous function of two variables is given f(x,y). Let's arbitrarily split the curve L on n parts with dots A = M 0, M 1, M 2, ... M n = B. Then on each of the resulting parts \(\bar((M)_(i-1)(M)_(i))\) we select any point \(\bar((M)_(i))\left(\ bar((x)_(i)),\bar((y)_(i))\right)\)and make the sum $$(S)_(n)=\sum_(i=1)^(n )f\left(\bar((x)_(i)),\bar((y)_(i))\right)\Delta (l)_(i)$$ where \(\Delta(l) _(i)=(M)_(i-1)(M)_(i)\) - arc of arc \(\bar((M)_(i-1)(M)_(i))\) . The amount received is called integral sum of the first kind for the function f(x,y) , given on the curve L.

    Let us denote by d the largest of the arc lengths \(\bar((M)_(i-1)(M)_(i))\) (thus d = \(max_(i)\Delta(l)_(i)\ )). If at d? 0 there is a limit of integral sums S n (independent of the method of partitioning the curve L into parts and the choice of points \(\bar((M)_(i))\)), then this limit is called first order curvilinear integral from function f(x,y) along the curve L and is denoted by $$\int_(L)f(x,y)dl$$

    It can be proven that if the function f(x,y) is continuous, then the line integral \(\int_(L)f(x,y)dl\) exists.

    Properties of a curvilinear integral of the 1st kind

    A curvilinear integral of the first kind has properties similar to the corresponding properties of a definite integral:

    • additivity,
    • linearity,
    • module assessment,
    • mean value theorem.

    However, there is a difference: $$\int_(AB)f(x,y)dl=\int_(BA)f(x,y)dl$$ i.e. a line integral of the first kind does not depend on the direction of integration.

    Calculation of curvilinear integrals of the first kind

    The calculation of a curvilinear integral of the first kind is reduced to the calculation of a definite integral. Namely:

    1. If the curve L is given by a continuously differentiable function y=y(x), x \(\in \) , then $$(\int\limits_L (f\left((x,y) \right)dl) ) = (\int \limits_a^b (f\left((x,y\left(x \right)) \right)\sqrt (1 + ((\left((y"\left(x \right)) \right))^ 2)) dx) ;)$$ in this case the expression \(dl=\sqrt((1 + ((\left((y"\left(x \right)) \right))^2))) dx \) is called the arc length differential.
    2. If the curve L is specified parametrically, i.e. in the form x=x(t), y=y(t), where x(t), y(t) are continuously differentiable functions on some interval \(\left [ \alpha ,\beta \right ]\), then $$ (\int\limits_L (f\left((x,y) \right)dl) ) = (\int\limits_\alpha ^\beta (f\left ((x\left(t \right),y \left(t \right)) \right)\sqrt (((\left((x"\left(t \right)) \right))^2) + ((\left((y"\left(t \right)) \right))^2)) dt)) $$ This equality extends to the case of a spatial curve L defined parametrically: x=x(t), y=y(t), z=z(t), \(t\in \left [ \alpha ,\beta \right ]\). In this case, if f(x,y,z) is a continuous function along the curve L, then $$ (\int\limits_L (f\left((x,y,z) \right)dl) ) = (\int \limits_\alpha ^\beta (f\left [ (x\left(t \right),y\left(t \right),z\left(t \right)) \right ]\sqrt (((\left ((x"\left(t \right)) \right))^2) + ((\left((y"\left(t \right)) \right))^2) + ((\left(( z"\left(t \right)) \right))^2)) dt)) $$
    3. If a plane curve L is given by the polar equation r=r(\(\varphi \)), \(\varphi \in\left [ \alpha ,\beta \right ] \), then $$ (\int\limits_L (f\ left((x,y) \right)dl) ) = (\int\limits_\alpha ^\beta (f\left((r\cos \varphi ,r\sin \varphi ) \right)\sqrt ((r ^2) + (((r)")^2)) d\varphi)) $$

    Curvilinear integrals of the 1st kind - examples

    Example 1

    Calculate a line integral of the first kind

    $$ \int_(L)\frac(x)(y)dl $$ where L is the arc of the parabola y 2 =2x, enclosed between points (2,2) and (8,4).

    Solution: Find the differential of the arc dl for the curve \(y=\sqrt(2x)\). We have:

    \((y)"=\frac(1)(\sqrt(2x)) \) $$ dl=\sqrt(1+\left ((y)" \right)^(2)) dx= \sqrt( 1+\left (\frac(1)(\sqrt(2x)) \right)^(2)) dx = \sqrt(1+ \frac(1)(2x)) dx $$ Therefore this integral is equal to: $ $\int_(L)\frac(x)(y)dl=\int_(2)^(8)\frac(x)(\sqrt(2x))\sqrt(1+\frac(1)(2x) )dx= \int_(2)^(8)\frac(x\sqrt(1+2x))(2x)dx= $$ $$ \frac(1)(2)\int_(2)^(8) \sqrt(1+2x)dx = \frac(1)(2).\frac(1)(3)\left (1+2x \right)^(\frac(3)(2))|_(2 )^(8)= \frac(1)(6)(17\sqrt(17)-5\sqrt(5)) $$

    Example 2

    Calculate the curvilinear integral of the first kind \(\int_(L)\sqrt(x^2+y^2)dl \), where L is the circle x 2 +y 2 =ax (a>0).

    Solution: Let's introduce polar coordinates: \(x = r\cos \varphi \), \(y=r\sin \varphi \). Then since x 2 +y 2 =r 2, the equation of the circle has the form: \(r^(2)=arcos\varphi \), that is, \(r=acos\varphi \), and the differential of the arc $$ dl = \ sqrt(r^2+(2)"^2)d\varphi = $$ $$ =\sqrt(a^2cos^2\varphi=a^2sin^2\varphi )d\varphi=ad\varphi $$ .

    In this case, \(\varphi\in \left [- \frac(\pi )(2) ,\frac(\pi )(2) \right ] \). Therefore, $$ \int_(L)\sqrt(x^2+y^2)dl=a\int_(-\frac(\pi )(2))^(\frac(\pi )(2))acos \varphi d\varphi =2a^2 $$

    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 is equal to the length of the arc AB .

    As always in integral calculus, a line 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 line 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 before 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 before 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.

    Department of Higher Mathematics

    Curvilinear integrals

    Guidelines

    Volgograd


    UDC 517.373(075)

    Reviewer:

    Senior Lecturer of the Department of Applied Mathematics N.I. Koltsova

    Published by decision of the editorial and publishing council

    Volgograd State Technical University

    Curvilinear integrals: method. instructions / comp. M.I. Andreeva,

    O.E. Grigorieva; Volga State Technical University. – Volgograd, 2011. – 26 p.

    The guidelines are a guide to completing individual assignments on the topic “Curvilinear integrals and their applications to field theory.”

    The first part of the guidelines contains the necessary theoretical material for completing individual tasks.

    The second part examines examples of performing all types of tasks included in individual assignments on the topic, which contributes to better organization of students’ independent work and successful mastery of the topic.

    The guidelines are intended for first and second year students.

    © Volgograd State

    Technical University, 2011

    1. CURVILINEAR INTEGRAL OF THE 1ST KIND

    Definition of a curvilinear integral of the 1st kind

    Let È AB– arc of a plane or spatial piecewise smooth curve L, f(P) is a continuous function defined on this arc, A 0 = A, A 1 , A 2 , …, A n – 1 , A n = B AB And P i– arbitrary points on partial arcs È A i – 1 A i, whose lengths are D l i (i = 1, 2, …, n

    at n® ¥ and max D l i® 0, which does not depend on the method of partitioning the arc È AB dots A i, nor from the choice of points P i on partial arcs È A i – 1 A i (i = 1, 2, …, n). This limit is called the curvilinear integral of the 1st kind of the function f(P) along the curve L and is designated

    Calculation of a curvilinear integral of the 1st kind

    The calculation of a curvilinear integral of the 1st kind can be reduced to the calculation of a definite integral using different methods of specifying the integration curve.

    If arc È AB plane curve is given parametrically by the equations where x(t) And y(t t, and x(t 1) = x A, x(t 2) = x B, That

    Where - differential of the arc length of the curve.

    A similar formula holds in the case of a parametric specification of a spatial curve L. If arc È AB crooked L is given by the equations , and x(t), y(t), z(t) – continuously differentiable functions of the parameter t, That

    where is the differential of the arc length of the curve.

    in Cartesian coordinates

    If arc È AB flat curve L given by the equation Where y(x

    and the formula for calculating the curvilinear integral is:

    When specifying an arc È AB flat curve L as x= x(y), y Î [ y 1 ; y 2 ],
    Where x(y) is a continuously differentiable function,

    and the curvilinear integral is calculated by the formula

    (1.4)

    Defining an Integration Curve by a Polar Equation

    If the curve is flat L is given by the equation in the polar coordinate system r = r(j), j О , where r(j) is a continuously differentiable function, then

    And

    (1.5)

    Applications of curvilinear integral of the 1st kind

    Using a curvilinear integral of the 1st kind, the following are calculated: the arc length of a curve, the area of ​​a part of a cylindrical surface, mass, static moments, moments of inertia and coordinates of the center of gravity of a material curve with a given linear density.

    1. Length l flat or spatial curve L is found by the formula

    2. Area of ​​a part of a cylindrical surface parallel to the axis OZ generatrix and located in the plane XOY guide L, enclosed between the plane XOY and the surface given by the equation z = f(x; y) (f(P) ³ 0 at P Î L), is equal to

    (1.7)

    3. Weight m material curve L with linear density m( P) is determined by the formula

    (1.8)

    4. Static moments about the axes Ox And Oy and coordinates of the center of gravity of a plane material curve L with linear density m( x; y) are respectively equal:

    (1.9)

    5. Static moments about planes Oxy, Oxz, Oyz and coordinates of the center of gravity of a spatial material curve with linear density m( x; y; z) are determined by the formulas:

    (1.11)

    6. For a flat material curve L with linear density m( x; y) moments of inertia about the axes Ox, Oy and the origin of coordinates are respectively equal:

    (1.13)

    7. Moments of inertia of a spatial material curve L with linear density m( x; y; z) relative to coordinate planes are calculated using the formulas

    (1.14)

    and the moments of inertia about the coordinate axes are equal to:

    (1.15)

    2. CURVILINEAR INTEGRAL OF THE 2nd KIND

    Definition of a curvilinear integral of the 2nd kind

    Let È AB– arc of a piecewise smooth oriented curve L, = (a x(P); a y(P); a z(P)) is a continuous vector function defined on this arc, A 0 = A, A 1 , A 2 , …, A n – 1 , A n = B– arbitrary arc split AB And P i– arbitrary points on partial arcs A i – 1 A i. Let be a vector with coordinates D x i, D y i, D z i(i = 1, 2, …, n), and is the scalar product of vectors and ( i = 1, 2, …, n). Then there is a limit of the sequence of integral sums

    at n® ¥ and max ÷ ç ® 0, which does not depend on the method of dividing the arc AB dots A i, nor from the choice of points P i on partial arcs È A i – 1 A i
    (i = 1, 2, …, n). This limit is called the curvilinear integral of the 2nd kind of the function ( P) along the curve L and is designated

    In the case when the vector function is specified on a plane curve L, similarly we have:

    When the direction of integration changes, the curvilinear integral of the 2nd kind changes sign.

    Curvilinear integrals of the first and second kind are related by the relation

    (2.2)

    where is the unit vector of the tangent to the oriented curve.

    Using a curvilinear integral of the 2nd kind, you can calculate the work of a force when moving a material point along the arc of a curve L:

    (2.3)

    Positive direction of traversing a closed curve WITH, bounding a simply connected region G, counterclockwise traversal is considered.

    Curvilinear integral of the 2nd kind over a closed curve WITH is called circulation and is denoted

    (2.4)

    Calculation of a curvilinear integral of the 2nd kind

    The calculation of a curvilinear integral of the 2nd kind is reduced to the calculation of a definite integral.

    Parametric definition of the integration curve

    If È AB oriented plane curve is given parametrically by the equations where X(t) And y(t) – continuously differentiable functions of the parameter t, and then

    (2.5)

    A similar formula takes place in the case of a parametric specification of a spatial oriented curve L. If arc È AB crooked L is given by the equations , and – continuously differentiable functions of the parameter t, That

    (2.6)

    Explicitly specifying a plane integration curve

    If arc È AB L is given in Cartesian coordinates by the equation where y(x) is a continuously differentiable function, then

    (2.7)

    When specifying an arc È AB plane oriented curve L as
    x= x(y), y Î [ y 1 ; y 2 ], where x(y) is a continuously differentiable function, the formula is valid

    (2.8)

    Let the functions are continuous along with their derivatives

    in a flat closed region G, bounded by a piecewise smooth closed self-disjoint positively oriented curve WITH+ . Then Green's formula holds:

    Let G– surface-simply connected region, and

    = (a x(P); a y(P); a z(P))

    – vector field specified in this area. Field ( P) is called potential if such a function exists U(P), What

    (P) = grad U(P),

    Necessary and sufficient condition for the potentiality of a vector field ( P) has the form:

    rot( P) = , where (2.10)

    (2.11)

    If the vector field is potential, then the curvilinear integral of the 2nd kind does not depend on the integration curve, but depends only on the coordinates of the beginning and end of the arc M 0 M. Potential U(M) of the vector field is determined up to a constant term and is found by the formula

    (2.12)

    Where M 0 M– an arbitrary curve connecting a fixed point M 0 and variable point M. To simplify calculations, a broken line can be chosen as the integration path M 0 M 1 M 2 M with links parallel to the coordinate axes, for example:

    3. examples of completing tasks

    Exercise 1

    Calculate a curvilinear integral of the first kind

    where L is the arc of the curve, 0 ≤ x ≤ 1.

    Solution. Using formula (1.3) to reduce a curvilinear integral of the first kind to a definite integral in the case of a smooth plane explicitly defined curve:

    Where y = y(x), x 0 ≤ x ≤ x 1 – arc equation L integration curve. In the example under consideration Find the derivative of this function

    and the arc length differential of the curve L

    ,

    then, substituting into this expression instead of y, we get

    Let us transform the curvilinear integral to a definite integral:

    We calculate this integral using substitution. Then
    t 2 = 1 + x, x = t 2 – 1, dx = 2t dt; at x = 0 t= 1; A x= 1 corresponds to . After transformations we get

    Task 2

    Calculate the curvilinear integral of the 1st kind along an arc L crooked L:x= cos 3 t, y= sin 3 t, .

    Solution. Because L is an arc of a smooth plane curve, given in parametric form, then we use formula (1.1) for reducing a curvilinear integral of the 1st kind to a definite one:

    .

    In the example under consideration

    Let's find the arc length differential

    We substitute the found expressions into formula (1.1) and calculate:

    Task 3

    Find the mass of the arc of the line L with linear plane m.

    Solution. Weight m arcs L with density m( P) is calculated using formula (1.8)

    .

    This is a curvilinear integral of the 1st kind over a parametrically defined smooth arc of a curve in space, therefore it is calculated using formula (1.2) for reducing a curvilinear integral of the 1st kind to a definite integral:

    Let's find derivatives

    and arc length differential

    We substitute these expressions into the formula for mass:

    Task 4

    Example 1. Calculate curvilinear integral of the 2nd kind

    along an arc L curve 4 x + y 2 = 4 from point A(1; 0) to point B(0; 2).

    Solution. Flat arc L is specified implicitly. To calculate the integral, it is more convenient to express x through y:

    and find the integral using formula (2.8) for transforming a curvilinear integral of the 2nd kind into a definite integral over a variable y:

    Where a x(x; y) = xy – 1, a y(x; y) = xy 2 .

    Taking into account the specification of the curve

    Using formula (2.8) we obtain

    Example 2. Calculate curvilinear integral of the 2nd kind

    Where L– broken line ABC, A(1; 2), B(3; 2), C(2; 1).

    Solution. By the property of additivity of a curvilinear integral

    Each of the integral terms is calculated using formula (2.7)

    Where a x(x; y) = x 2 + y, a y(x; y) = –3xy.

    Equation of a line segment AB: y = 2, y¢ = 0, x 1 = 1, x 2 = 3. Substituting these expressions into formula (2.7), we obtain:

    To calculate the integral

    let's make an equation of a straight line B.C. according to the formula

    Where x B, y B, x C, y C– coordinates of points B And WITH. We get

    y – 2 = x – 3, y = x – 1, y¢ = 1.

    We substitute the resulting expressions into formula (2.7):

    Task 5

    Calculate a curvilinear integral of the 2nd kind along an arc L

    0 ≤ t ≤ 1.

    Solution. Since the integration curve is given parametrically by the equations x = x(t), y = y(t), t Î [ t 1 ; t 2 ], where x(t) And y(t) – continuously differentiable functions t at t Î [ t 1 ; t 2 ], then to calculate the curvilinear integral of the second kind we use formula (2.5) reducing the curvilinear integral to the one defined for a plane parametrically given curve

    In the example under consideration a x(x; y) = y; a y(x; y) = –2x.

    Taking into account the curve setting L we get:

    We substitute the found expressions into formula (2.5) and calculate the definite integral:

    Task 6

    Example 1. C + Where WITH : y 2 = 2x, y = x – 4.

    Solution. Designation C+ indicates that the circuit is traversed in the positive direction, that is, counterclockwise.

    Let us check that to solve the problem we can use Green’s formula (2.9)

    Since the functions a x (x; y) = 2y – x 2 ; a y (x; y) = 3x + y and their partial derivatives continuous in a flat closed region G, limited by contour C, then Green's formula is applicable.

    To calculate the double integral, we depict the region G, having previously determined the intersection points of the arcs of curves y 2 = 2x And
    y = x– 4, making up the contour C.

    We will find the intersection points by solving the system of equations:

    The second equation of the system is equivalent to the equation x 2 – 10x+ 16 = 0, whence x 1 = 2, x 2 = 8, y 1 = –2, y 2 = 4.

    So, the points of intersection of the curves: A(2; –2), B(8; 4).

    Since the area G– correct in the direction of the axis Ox, then to reduce the double integral to a repeated one, we project the region G per axis OY and use the formula

    .

    Because a = –2, b = 4, x 2 (y) = 4+y, That

    Example 2. Calculate a curvilinear integral of the 2nd kind along a closed contour Where WITH– outline of a triangle with vertices A(0; 0), B(1; 2), C(3; 1).

    Solution. The designation means that the contour of the triangle is traversed clockwise. In the case where the curvilinear integral is taken over a closed contour, Green's formula takes the form

    Let's depict the area G, limited by a given contour.

    Functions and partial derivatives And continuous in the area G, so Green's formula can be applied. Then

    Region G is not correct in the direction of any of the axes. Let's draw a straight line segment x= 1 and imagine G as G = G 1 È G 2 where G 1 and G 2 areas correct in axis direction Oy.

    Then

    To reduce each of the double integrals by G 1 and G 2 to repeat we will use the formula

    Where [ a; b] – area projection D per axis Ox,

    y = y 1 (x) – equation of the lower bounding curve,

    y = y 2 (x) – equation of the upper limiting curve.

    Let us write down the equations of the domain boundaries G 1 and find

    AB: y = 2x, 0 ≤ x ≤ 1; AD: , 0 ≤ x ≤ 1.

    Let's create an equation for the boundary B.C. region G 2 using the formula

    B.C.: where 1 ≤ x ≤ 3.

    DC: 1 ≤ x ≤ 3.

    Task 7

    Example 1. Find the work of force L: y = x 3 from point M(0; 0) to point N(1; 1).

    Solution. Work done by a variable force when moving a material point along an arc of a curve L determined by formula (2.3) (as a curvilinear integral of the second kind of a function along the curve L) .

    Since the vector function is given by the equation and the arc of the plane oriented curve is defined explicitly by the equation y = y(x), x Î [ x 1 ; x 2 ], where y(x) is a continuously differentiable function, then by formula (2.7)

    In the example under consideration y = x 3 , , x 1 = x M = 0, x 2 = x N= 1. Therefore

    Example 2. Find the work of force when moving a material point along a line L: x 2 + y 2 = 4 from point M(0; 2) to point N(–2; 0).

    Solution. Using formula (2.3), we obtain

    .

    In the example under consideration, the arc of the curve L(È MN) is a quarter of a circle given by the canonical equation x 2 + y 2 = 4.

    To calculate a curvilinear integral of the second kind, it is more convenient to go to the parametric definition of a circle: x = R cos t, y = R sin t and use formula (2.5)

    Because x= 2cos t, y= 2sin t, , , we get

    Task 8

    Example 1. Calculate the circulation modulus of a vector field along the contour G:

    Solution. To calculate the circulation of a vector field along a closed contour G let's use formula (2.4)

    Since the spatial vector field is given and spatial closed loop G, then passing from the vector form of writing the curvilinear integral to the coordinate form, we obtain

    Curve G defined as the intersection of two surfaces: a hyperbolic paraboloid z = x 2 – y 2 + 2 and cylinders x 2 + y 2 = 1. To calculate the curvilinear integral it is convenient to go to the parametric equations of the curve G.

    The equation of a cylindrical surface can be written as:
    x=cos t, y= sin t, z = z. Expression for z in the parametric equations of the curve is obtained by substituting x=cos t, y= sin t into the equation of a hyperbolic paraboloid z = 2 + cos 2 t– sin 2 t= 2 + cos 2 t. So, G: x=cos t,
    y= sin t, z= 2 + cos 2 t, 0 ≤ t≤ 2p.

    Since those included in the parametric equations of the curve G functions
    x(t) = cos t, y(t) = sin t, z(t) = 2 + cos 2 t are continuously differentiable functions of the parameter t at tО , then we find the curvilinear integral using formula (2.6)