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, we 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 the 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

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, $$ \int_(L)\sqrt(x^2+y^2)dl=a\int_(-\frac(\pi )(2))^(\frac(\pi )(2))acos \varphi d\varphi =2a^2 $$

A curvilinear integral of the 2nd kind is calculated in the same way as a curvilinear integral of the 1st kind by reduction to the definite. To do this, all variables under the integral sign are expressed through one variable, using the equation of the line along which the integration is performed.

a) If the line AB is given by a system of equations then

(10.3)

For the plane case, when the curve is given by the equation the curvilinear integral is calculated using the formula: . (10.4)

If the line AB is given by parametric equations then

(10.5)

For a flat case, if the line AB given by parametric equations , the curvilinear integral is calculated by the formula:

, (10.6)

where are the parameter values t, corresponding to the starting and ending points of the integration path.

If the line AB piecewise smooth, then we should use the property of additivity of the curvilinear integral by splitting AB on smooth arcs.

Example 10.1 Let's calculate the curvilinear integral along a contour consisting of part of a curve from a point before and ellipse arcs from point before .

Since the contour consists of two parts, we use the additivity property of the curvilinear integral: . Let us reduce both integrals to definite ones. Part of the contour is given by an equation relative to the variable . Let's use the formula (10.4 ), in which we switch the roles of the variables. Those.

. After calculation we get .

To calculate the contour integral Sun Let's move on to the parametric form of writing the ellipse equation and use formula (10.6).

Pay attention to the limits of integration. Point corresponds to the value, and to the point corresponds Answer:
.

Example 10.2. Let's calculate along a straight line segment AB, Where A(1,2,3), B(2,5,8).

Solution. A curvilinear integral of the 2nd kind is given. To calculate it, you need to convert it to a specific one. Let's compose the equations of the line. Its direction vector has coordinates .

Canonical equations of line AB: .

Parametric equations of this line: ,

At
.

Let's use the formula (10.5) :

Having calculated the integral, we get the answer: .

5. Work of force when moving a material point of unit mass from point to point along a curve .

Let at each point of a piecewise smooth curve a vector is given that has continuous coordinate functions: . Let's break this curve into small parts with points so that at the points of each part meaning of functions
could be considered constant, and the part itself could be mistaken for a straight segment (see Fig. 10.1). Then . The scalar product of a constant force, the role of which is played by a vector , per rectilinear displacement vector is numerically equal to the work done by the force when moving a material point along . Let's make an integral sum . In the limit, with an unlimited increase in the number of partitions, we obtain a curvilinear integral of the 2nd kind

. (10.7) Thus, the physical meaning of the curvilinear integral of the 2nd kind - this is work done by force when moving a material point from A To IN along the contour L.

Example 10.3. Let's calculate the work done by the vector when moving a point along a portion of a Viviani curve defined as the intersection of a hemisphere and cylinder , running counterclockwise when viewed from the positive part of the axis OX.

Solution. Let's construct the given curve as the line of intersection of two surfaces (see Fig. 10.3).

.

To reduce the integrand to one variable, let’s move to a cylindrical coordinate system: .

Because a point moves along a curve , then it is convenient to choose as a parameter a variable that changes along the contour so that . Then we obtain the following parametric equations of this curve:

.Wherein
.

Let us substitute the resulting expressions into the formula for calculating circulation:

( - the + sign indicates that the point moves along the contour counterclockwise)

Let's calculate the integral and get the answer: .

Lesson 11.

Green's formula for a simply connected region. Independence of the curvilinear integral from the path of integration. Newton-Leibniz formula. Finding a function from its total differential using a curvilinear integral (plane and spatial cases).

OL-1 chapter 5, OL-2 chapter 3, OL-4 chapter 3 § 10, clause 10.3, 10.4.

Practice : OL-6 No. 2318 (a, b, d), 2319 (a, c), 2322 (a, d), 2327, 2329 or OL-5 No. 10.79, 82, 133, 135, 139.

Home building for lesson 11: OL-6 No. 2318 (c, d), 2319 (c, d), 2322 (b, c), 2328, 2330 or OL-5 No. 10.80, 134, 136, 140

Green's formula.

Let on the plane given a simply connected domain bounded by a piecewise smooth closed contour. (A region is called simply connected if any closed contour in it can be contracted to a point in this region).

Theorem. If the functions and their partial derivatives G, That

Figure 11.1

- Green's formula . (11.1)

Indicates positive bypass direction (counterclockwise).

Example 11.1. Using Green's formula, we calculate the integral along a contour consisting of segments O.A., O.B. and greater arc of a circle , connecting the points A And B, If , , .

Solution. Let's build a contour (see Fig. 11.2). Let us calculate the necessary derivatives.

Figure 11.2

, ; , . Functions and their derivatives are continuous in a closed region bounded by a given contour. According to Green's formula, this integral is .

After substituting the calculated derivatives we get

. We calculate the double integral by moving to polar coordinates:
.

Let's check the answer by calculating the integral directly along the contour as a curvilinear integral of the 2nd kind.
.

Answer:
.

2. Independence of the curvilinear integral from the path of integration.

Let And - arbitrary points of a simply connected region pl. . Curvilinear integrals calculated from different curves connecting these points generally have different meanings. But if certain conditions are met, all these values ​​may turn out to be the same. Then the integral does not depend on the shape of the path, but depends only on the starting and ending points.

The following theorems hold.

Theorem 1. In order for the integral
did not depend on the shape of the path connecting the points and , it is necessary and sufficient that this integral along any closed contour be equal to zero.

Theorem 2.. In order for the integral
along any closed contour is equal to zero, it is necessary and sufficient that the function and their partial derivatives were continuous in a closed region G and so that the condition ( 11.2)

Thus, if the conditions for the integral to be independent of the path shape are met (11.2) , then it is enough to specify only the start and end points: (11.3)

Theorem 3. If the condition is satisfied in a simply connected region, then there is a function such that . (11.4)

This formula is called formula Newton–Leibniz for the line integral.

Comment. Recall that equality is a necessary and sufficient condition for the fact that the expression
.

Then from the above theorems it follows that if the functions and their partial derivatives continuous in a closed region G, in which the points are given And , and , then

a) there is a function , such that ,

does not depend on the shape of the path, ,

c) the formula holds Newton–Leibniz .

Example 11.2. Let us make sure that the integral
does not depend on the shape of the path, and let's calculate it.

Solution. .

Figure 11.3

Let's check that condition (11.2) is satisfied.
. As we can see, the condition is met. The value of the integral does not depend on the path of integration. Let us choose the integration path. Most

a simple way to calculate is a broken line DIA, connecting the starting and ending points of the path. (See Fig. 11.3)

Then .

3. Finding a function by its total differential.

Using a curvilinear integral, which does not depend on the shape of the path, we can find the function , knowing its full differential. This problem is solved as follows.

If the functions and their partial derivatives continuous in a closed region G and , then the expression is the total differential of some function . In addition, the integral
, firstly, does not depend on the shape of the path and, secondly, can be calculated using the Newton–Leibniz formula.

Let's calculate
two ways.

Figure 11.4

a) Select a point in the region with specific coordinates and a point with arbitrary coordinates. Let us calculate the curvilinear integral along a broken line consisting of two line segments connecting these points, with one of the segments parallel to the axis and the other to the axis. Then . (See Fig. 11.4)

The equation .

The equation .

We get: Having calculated both integrals, we get some function in the answer.

b) Now we calculate the same integral using the Newton–Leibniz formula.

Now let's compare two results of calculating the same integral. The functional part of the answer in point a) is the required function , and the numerical part is its value at the point .

Example 11.3. Let's make sure that the expression
is the total differential of some function and we'll find her. Let's check the results of calculating example 11.2 using the Newton-Leibniz formula.

Solution. Condition for the existence of a function (11.2) was checked in the previous example. Let's find this function, for which we will use Figure 11.4, and take for point . Let's compose and calculate the integral along the broken line DIA, Where :

As mentioned above, the functional part of the resulting expression is the desired function
.

Let's check the result of the calculations from Example 11.2 using the Newton–Leibniz formula:

The results were the same.

Comment. All the statements considered are also true for the spatial case, but with a larger number of conditions.

Let a piecewise smooth curve belong to a region in space . Then, if the functions and their partial derivatives are continuous in the closed region in which the points are given and , and
(11.5 ), That

a) the expression is the total differential of some function ,

b) curvilinear integral of the total differential of some function does not depend on the shape of the path and ,

c) the formula holds Newton–Leibniz .(11.6 )

Example 11.4. Let's make sure that the expression is the complete differential of some function and we'll find her.

Solution. To answer the question of whether a given expression is a complete differential of some function , let's calculate the partial derivatives of the functions, , . (Cm. (11.5) ) ; ; ; ; ; .

These functions are continuous along with their partial derivatives at any point in space.

We see that the necessary and sufficient conditions for existence are satisfied : , , , etc.

To calculate a function Let us take advantage of the fact that the linear integral does not depend on the path of integration and can be calculated using the Newton-Leibniz formula. Let the point - the beginning of the path, and some point - end of the road . Let's calculate the integral

along a contour consisting of straight segments parallel to the coordinate axes. (see Fig. 11.5).

.

Figure 11.5

Equations of the contour parts: , ,
.

Then

, x fixed here, so ,

Recorded here y, That's why .

As a result we get: .

Now let's calculate the same integral using the Newton-Leibniz formula.

Let's compare the results: .

From the resulting equality it follows that , and

Lesson 12.

Surface integral of the first kind: definition, basic properties. Rules for calculating a surface integral of the first kind using a double integral. Applications of the surface integral of the first kind: surface area, mass of a material surface, static moments about coordinate planes, moments of inertia, and coordinates of the center of gravity. OL-1 ch.6, OL 2 ch.3, OL-4§ 11.

Practice: OL-6 No. 2347, 2352, 2353 or OL-5 No. 10.62, 65, 67.

Homework for lesson 12:

OL-6 No. 2348, 2354 or OL-5 No. 10.63, 64, 68.

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 the 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.

It is more convenient to calculate volume in cylindrical coordinates. Equation of a circle bounding a region D, a cone and a paraboloid

respectively take the form ρ = 2, z = ρ, z = 6 − ρ 2. Taking into account the fact that this body is symmetrical relative to the xOz and yOz planes. we have

	6− ρ 2
V = 4 ∫ 2 dϕ ∫ ρ dρ ∫ dz = 4 ∫ 2 dϕ ∫ ρ z				6 ρ − ρ 2 d ρ =

4 ∫ d ϕ∫ (6 ρ − ρ3 − ρ2 ) d ρ =

2 d ϕ =

4 ∫ 2 (3 ρ 2 −

∫ 2 d ϕ =

32π

If symmetry is not taken into account, then

6− ρ 2

32π

V = ∫

dϕ ∫ ρ dρ ∫ dz =

3. CURVILINEAR INTEGRALS

Let us generalize the concept of a definite integral to the case when the domain of integration is a certain curve. Integrals of this kind are called curvilinear. There are two types of curvilinear integrals: curvilinear integrals along the length of the arc and curvilinear integrals over the coordinates.

3.1. Definition of a curvilinear integral of the first type (along the length of the arc). Let the function f(x,y) defined along a flat piecewise

smooth1 curve L, the ends of which will be points A and B. Let us divide the curve L arbitrarily into n parts with points M 0 = A, M 1,... M n = B. On

For each of the partial arcs M i M i + 1, we select an arbitrary point (x i, y i) and calculate the values ​​of the function f (x, y) at each of these points. Sum

1 A curve is called smooth if at each point there is a tangent that continuously changes along the curve. A piecewise smooth curve is a curve consisting of a finite number of smooth pieces.

n− 1
σ n = ∑ f (x i , y i ) ∆ l i ,

i = 0

where ∆ l i is the length of the partial arc M i M i + 1, called integral sum

for the function f(x, y) along the curve L. Let us denote the largest of the lengths
partial arcs M i M i + 1 , i =
	0 ,n − 1 through λ , that is, λ = max ∆ l i .
		0 ≤i ≤n −1
If there is a finite limit I of the integral sum (3.1)
tending to zero of the largest of the lengths of partial arcsM i M i + 1,
depending neither on the method of dividing the curve L into partial arcs, nor on

choice of points (x i, y i), then this limit is called curvilinear integral of the first type (curvilinear integral along the length of the arc) from the function f (x, y) along the curve L and is denoted by the symbol ∫ f (x, y) dl.

Thus, by definition
n− 1
I = lim ∑ f (xi , yi ) ∆ li = ∫ f (x, y) dl.
λ → 0 i = 0

The function f(x, y) is called in this case integrable along the curve L,

the curve L = AB is the contour of integration, A is the initial point, and B is the final point of integration, dl is the element of arc length.

Remark 3.1. If in (3.2) we put f (x, y) ≡ 1 for (x, y) L, then

we obtain an expression for the length of the arc L in the form of a curvilinear integral of the first type

l = ∫ dl.

Indeed, from the definition of a curvilinear integral it follows that
	dl = lim n − 1
		∆l	Lim l = l .
	λ → 0 ∑		λ→ 0
	i = 0
3.2. Basic properties of the first type of curvilinear integral
are similar to the properties of a definite integral:
1 o. ∫ [ f1 (x, y) ± f2 (x, y) ] dl = ∫ f1 (x, y) dl ± ∫ f2 (x, y) dl.
2 o. ∫ cf (x, y) dl = c ∫ f (x, y) dl, where c is a constant.
				and L, not
3 o. If the integration loop L is divided into two parts L

having common interior points, then

∫ f (x, y)dl = ∫ f (x, y)dl + ∫ f (x, y)dl.

4 o. We especially note that the value of the curvilinear integral of the first type does not depend on the direction of integration, since the values ​​of the function f (x, y) in

arbitrary points and the length of partial arcs ∆ l i , which are positive,

regardless of which point of the curve AB is considered the initial and which is the final, that is

	f (x, y) dl = ∫ f (x, y) dl .
3.3. Calculation of a curve integral of the first type
reduces to calculating definite integrals.
	x= x(t)
Let the curve L given by parametric equations		y=y(t)

Let α and β be the values ​​of the parameter t corresponding to the beginning (point A) and

end (point B)

[α , β ]

x(t), y(t) and

derivatives

x (t), y (t)

Continuous

f(x, y) -

is continuous along the curve L. From the course of differential calculus

functions of one variable it is known that

dl = (x(t))

+ (y(t))

∫ f (x, y) dl = ∫ f (x(t), y(t))

(x(t)

+ (y(t))

∫ x2 dl,

Example 3.1.

Calculate

circle

x= a cos t

0 ≤ t ≤

y= a sin t

Solution. Since x (t) = − a sin t, y (t) = a cos t, then

dl =

(− a sin t) 2 + (a cos t) 2 dt = a2 sin 2 t + cos 2 tdt = adt

and from formula (3.4) we obtain

Cos 2t )dt =

sin 2t

∫ x2 dl = ∫ a2 cos 2 t adt = a

3 ∫

πa 3

sinπ

L is given

equation

y = y(x) ,

a ≤ x ≤ b

y(x)

is continuous along with its derivative y

(x) for a ≤ x ≤ b, then

dl =

1+(y(x))

and formula (3.4) takes the form

∫ f (x, y) dl = ∫ f (x, y(x))

(y(x))

L is given

x = x(y), c ≤ y ≤ d

x(y)

equation

is continuous along with its derivative x (y) for c ≤ y ≤ d, then

dl =

1+(x(y))

and formula (3.4) takes the form

∫ f (x, y) dl = ∫ f (x(y), y)

1 + (x(y))

Example 3.2. Calculate ∫ ydl, where L is the arc of the parabola

2 x from

point A (0,0) to point B (2,2).

Solution . Let's calculate the integral in two ways, using

formulas (3.5) and (3.6)

1) Let's use formula (3.5). Because

2x (y ≥ 0), y ′

2 x =

2 x

dl =

1+ 2 x dx,

3 / 2 2

1 (5

3 2 − 1) .

∫ ydl = ∫

2 x + 1 dx = ∫ (2 x + 1) 1/ 2 dx =

1 (2x + 1)

2) Let's use formula (3.6). Because

x = 2 , x

Y, dl

1 + y

		y 1 + y 2 dy =		(1 + y				/ 2 2
∫ ydl = ∫
				3 / 2

1 3 (5 5 − 1).

Remark 3.2. Similar to what was considered, we can introduce the concept of a curvilinear integral of the first type of function f (x, y, z) over

spatial piecewise smooth curve L:

If the curve L is given by parametric equations

α ≤ t ≤ β, then

dl =

(x(t))

(y(t))

(z(t))

∫ f (x, y, z) dl =

= ∫

dt.

f (x (t), y (t), z (t)) (x (t))

(y(t))

(z(t))

x= x(t) , y= y(t)

z= z(t)

Example 3.3. Calculate∫ (2 z − x 2 + y 2 ) dl , where L is the arc of the curve

x= t cos t	0 ≤ t ≤ 2 π.
y = t sin t
z = t
	x′ = cost − t sint, y′ = sint + t cost, z′ = 1 ,
	dl =	(cos t − t sin t)2 + (sin t + t cos t)2 + 1 dt =

Cos2 t − 2 t sin t cos t + t2 sin2 t + sin2 t + 2 t sin t cos t + t2 cos2 t + 1 dt =

2 + t2 dt .

Now, according to formula (3.7), we have

∫ (2z −

x2 + y2 ) dl = ∫ (2 t −

t 2 cos 2 t + t 2 sin 2 t )

2 + t 2 dt =

T2)

= ∫

t2+t

dt =

4π

− 2 2

cylindrical

surfaces,

which is made up of perpendiculars to

xOy plane,

restored at points

(x, y)

L=AB

and having

represents the mass of a curve L having a variable linear density ρ(x, y)

the linear density of which varies according to the law ρ (x, y) = 2 y.

Solution. To calculate the mass of the arc AB, we use formula (3.8). The arc AB is given parametrically, so to calculate the integral (3.8) we use formula (3.4). Because

1+t

dt,

x (t) = 1, y (t) = t, dl =

3/ 2 1

1 (1+ t

m = ∫ 2 ydl = ∫

1 2 + t2 dt = ∫ t 1 + t2 dt =

(2 3 / 2 −

1) =

2 2 − 1.

3.4. Definition of a curvilinear integral of the second type (by

coordinates). Let the function

f(x, y) is defined along a plane

piecewise smooth curve L, the ends of which will be points A and B. Again

arbitrary

let's break it

curve L

M 0 = A , M 1 ,... M n = B We also choose within

each partial

arcs M i M i + 1

arbitrary point

(xi, yi)

and calculate

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