Calculation procedure triple integral a is similar to the corresponding operation for a double integral. To describe it, we introduce the concept of a regular three-dimensional region:

Definition 9.1. A three-dimensional region V bounded by a closed surface S is called regular if:

1. any straight line parallel to the axis Oz and drawn through the interior point of the region intersects S at two points;
2. the entire region V is projected onto the Oxy plane into a regular two-dimensional region D;
3. any part of the region V, cut off from it by a plane parallel to any of the coordinate planes, has properties 1) and 2).

Let us consider a regular region V, bounded below and above by the surfaces z=χ(x,y) and z=ψ(x,y) and projected onto the Oxy plane into the regular region D, inside which x varies from a to b, limited by the curves y=φ1(x) and y=φ2(x) (Fig. 1). Let us define a continuous function f(x, y, z) in domain V.

Definition 9.2. Let us call the triple integral of the function f(x, y, z) over the region V an expression of the form:

The triple integral has the same properties as the double integral. We list them without proof, since they are proved similarly to the case of a double integral.

Calculation of the triple integral.

Theorem 9.1. The triple integral of the function f(x,y,z) over a regular domain V is equal to the triple integral over the same domain:

. (9.3)

Proof.

Let us divide the region V by planes parallel to the coordinate planes into n regular regions. Then from property 1 it follows that

where is the triple integral of the function f(x,y,z) over the region.

Using formula (9.2), the previous equality can be rewritten as:

From the condition of continuity of the function f(x,y,z) it follows that the limit of the integral sum on the right side of this equality exists and is equal to the triple integral. Then, passing to the limit at , we obtain:

Q.E.D.

Comment.

Similar to the case of a double integral, it can be proven that changing the order of integration does not change the value of the triple integral.

Example. Let us calculate the integral where V is a triangular pyramid with vertices at points (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1). Its projection onto the Oxy plane is a triangle with vertices (0, 0), (1, 0) and (0, 1). The region is limited from below by the plane z = 0, and from above by the plane x + y + z = 1. Let’s move on to the threefold integral:

Factors that do not depend on the integration variable can be taken out of the sign of the corresponding integral:

Curvilinear coordinate systems in three-dimensional space.

1. Cylindrical coordinate system.

The cylindrical coordinates of the point P(ρ,φ,z) are the polar coordinates ρ, φ of the projection of this point onto the Oxy plane and the applicate of this point z (Fig. 2).

Formulas for the transition from cylindrical to Cartesian coordinates can be specified as follows:

x = ρ cosφ, y = ρ sinφ, z = z. (9.4)

1. Spherical coordinate system.

In spherical coordinates, the position of a point in space is determined by the linear coordinate ρ - the distance from the point to the origin of the Cartesian coordinate system (or the pole of the spherical system), φ - the polar angle between the positive semi-axis Ox and the projection of the point onto the Oxy plane, and θ - the angle between the positive semi-axis of the axis Oz and segment OP (Fig. 3). Wherein

Let us set the formulas for the transition from spherical to Cartesian coordinates:

x = ρ sinθ cosφ, y = ρ sinθ sinφ, z = ρ cosθ. (9.5)

Jacobian and its geometric meaning.

Let us consider the general case of changing variables in a double integral. Let a region D be given in the Oxy plane, bounded by a line L. Let us assume that x and y are single-valued and continuously differentiable functions of new variables u and v:

x = φ(u, v), y = ψ(u, v). (9.6)

Let us consider the rectangular coordinate system Ouv, the point P΄(u, v) of which corresponds to the point P(x, y) from the region D. All such points form a region D΄ in the Ouv plane, bounded by the line L΄. We can say that formulas (9.6) establish a one-to-one correspondence between the points of the regions D and D΄. In this case, the lines u = const and

v = const in the Ouv plane will correspond to some lines in the Oxy plane.

Let us consider a rectangular area ΔS΄ in the Ouv plane, bounded by the straight lines u = const, u+Δu = const, v = const and v+Δv = const. It will correspond to a curved area ΔS in the Oxy plane (Fig. 4). The areas of the areas under consideration will also be denoted by ΔS΄ and ΔS. In this case, ΔS΄ = Δu Δv. Let's find the area ΔS. Let us denote the vertices of this curvilinear quadrilateral P1, P2, P3, P4, where

P1(x1, y1), x1 = φ(u, v), y1 = ψ(u, v);

P2(x2, y2), x2 = φ(u+Δu, v), y2 = ψ(u+Δu, v);

P3(x3, y3), x3 = φ(u+Δu, v+Δv), y3 = ψ(u+Δu, v+Δv);

P4(x4, y4), x4 = φ(u, v+Δv), y4 = ψ(u, v+Δv).

Let us replace the small increments Δu and Δv with the corresponding differentials. Then

In this case, the quadrilateral P1 P2 P3 P4 can be considered a parallelogram and its area can be determined using the formula from analytical geometry:

(9.7)

Definition 9.3. The determinant is called the functional determinant or Jacobian of the functions φ(x, y) and ψ(x, y).

Passing to the limit at in equality (9.7), we obtain the geometric meaning of the Jacobian:

that is, the module of the Jacobian is the limit of the ratio of the areas of infinitesimal areas ΔS and ΔS΄.

Comment. In a similar way, we can define the concept of a Jacobian and its geometric meaning for an n-dimensional space: if x1 = φ1(u1, u2,…,un), x2 = φ2(u1, u2,…,un),…, xn = φ(u1 , u2,…, un), then

(9.8)

In this case, the module of the Jacobian gives a limit to the ratio of the “volumes” of small regions of the spaces x1, x2,..., xn and u1, u2,..., un.

Change of variables in multiple integrals.

Let us study the general case of change of variables using the example of a double integral.

Let a continuous function z = f(x,y) be given in domain D, each value of which corresponds to the same value of the function z = F(u, v) in domain D΄, where

F(u, v) = f(φ(u, v), ψ(u, v)). (9.9)

Consider the integral sum

where the integral sum on the right is taken over the domain D΄. Passing to the limit at , we obtain a formula for transforming coordinates in the double integral.

Let us have two rectangular coordinate systems in space and
, and a system of functions

(1)

which establish a one-to-one correspondence between points in some areas
And
in these coordinate systems. Let us assume that the functions of system (1) have
continuous partial derivatives. The determinant made up of these partial derivatives

,

is called the Jacobian (or Jacobi determinant) of the system of functions (1). We will assume that
V
.

Under the assumptions made above, the following general formula for changing variables in a triple integral holds:

As in the case of the double integral, the mutual uniqueness of system (1) and the condition
may be violated at individual points, on individual lines and on individual surfaces.

System of functions (1) for each point
matches a single point
. These three numbers
are called curvilinear coordinates of a point . Points of space
, for which one of these coordinates retains a constant value, form the so-called. coordinate surface.

II Triple integral in cylindrical coordinates

The cylindrical coordinate system (CSS) is determined by the plane
, in which a polar coordinate system is specified and the axis
, perpendicular to this plane. Cylindrical coordinates of a point
, Where
– polar coordinates of the point – projections t glasses to the plane
, A – these are the coordinates of the projection of the point per axis
or
.

In plane
we enter Cartesian coordinates in the usual way, direct the applicate axis along the axis
CSK. Now it is not difficult to obtain formulas connecting cylindrical coordinates with Cartesian ones:

(3)

These formulas map the area to the entire space
.

The coordinate surfaces in the case under consideration will be:

1)
– cylindrical surfaces with generatrices parallel to the axis
, whose guides are circles in the plane
, centered at point ;

2)

;

3)
– planes parallel to the plane
.

Jacobian of system (3):

.

The general formula in the case of CSK takes the form:

Note 1 . The transition to cylindrical coordinates is recommended in the case when the area of ​​integration is a circular cylinder or cone, or a paraboloid of revolution (or parts thereof), and the axis of this body coincides with the axis of the applicate
.

Note 2. Cylindrical coordinates can be generalized in the same way as polar coordinates in a plane.

Example 1. Calculate the triple integral of a function

by region
, representing the inner part of the cylinder
, bounded by a cone
and paraboloid
.

Solution. We have already considered this area in §2, example 6, and obtained a standard entry in the DPSC. However, calculating the integral in this region is difficult. Let's go to CSK:

.

Projection
body
to the plane
- it's a circle
. Therefore, the coordinate varies from 0 to
, A – from 0 to R. Through an arbitrary point
draw a straight line parallel to the axis
. The straight line will go into
on a cone, but will come out on a paraboloid. But the cone
has the equation in the CSC
, and the paraboloid
- the equation
. So we have

III Triple integral in spherical coordinates

The spherical coordinate system (SCS) is determined by the plane
, in which the UCS is specified, and the axis
, perpendicular to the plane
.

Spherical coordinates of a point space is called a triple of numbers
, Where – polar angle of projection of a point onto a plane
,– angle between axis
and vector
And
.

In plane
let's introduce Cartesian coordinate axes
And
in the usual way, and the applicate axis is compatible with the axis
. The formulas connecting spherical coordinates with Cartesian ones are as follows:

(4)

These formulas map the area to the entire space
.

Jacobian of the system of functions (4):

.

There are three families of coordinate surfaces:

1)
– concentric spheres with the center at the origin;

2)
– half-planes passing through the axis
;

3)
– circular cones with a vertex at the origin of coordinates, the axis of which is the axis
.

Formula for transition to SSC in triple integral:

Note 3. The transition to the SCS is recommended when the domain of integration is a ball or part of it. In this case, the equation of the sphere
goes into. Like the CSK discussed earlier, the CSK is “tied” to the axis
. If the center of the sphere is shifted by a radius along the coordinate axis, then we obtain the simplest spherical equation when displaced along the axis
:

Note 4. It is possible to generalize the SSC:

with Jacobian
. This system of functions will translate the ellipsoid

to "parallelepiped"

Example 2. Find the average distance of points on a ball of radius from its center.

Solution. Recall that the average value of the function
in area
is the triple integral of a function over a region divided by the volume of the region. In our case

So we have

Triple integrals. Calculation of body volume.
Triple integral in cylindrical coordinates

For three days the dead man lay in the dean's office, dressed in Pythagoras' trousers,
In the hands of Fichtenholtz he held a volume that had brought him from this world,
A triple integral was tied to the legs, and the corpse was wrapped in a matrix,
And instead of praying, some impudent person read Bernoulli’s theorem.

Triple integrals are something that you don’t have to be afraid of =) Because if you are reading this text, then, most likely, you have a good understanding of theory and practice of “ordinary” integrals, and double integrals. And where there is a double, nearby there is a triple:

And really, what is there to fear? The integral is less, the integral is more....

Let's look at the recording:

– triple integral icon;
– integrand function of three variables;
– product of differentials.
– area of ​​integration.

Let us especially focus on areas of integration. If in double integral it represents flat figure, then here – spatial body, which, as is known, is limited by the set surfaces. Thus, in addition to the above, you must navigate basic surfaces of space and be able to make simple three-dimensional drawings.

Some are depressed, I understand... Alas, the article cannot be titled “triple integrals for dummies,” and there are some things you need to know/be able to do. But it’s okay - all the material is presented in an extremely accessible form and can be mastered in the shortest possible time!

What does it mean to calculate a triple integral and what is it even?

To calculate the triple integral means find the NUMBER:

In the simplest case, when the triple integral is numerically equal to the volume of the body. And indeed, according to general meaning of integration, the product is equal infinitesimal the volume of an elementary “brick” of the body. And the triple integral is just unites all these infinitesimal particles over the area, resulting in the integral (total) value of the volume of the body: .

In addition, the triple integral has important physical applications. But more about this later - in the 2nd part of the lesson, dedicated to calculations of arbitrary triple integrals, for which the function in the general case is different from a constant and is continuous in the region. In this article, we will consider in detail the problem of finding volume, which, according to my subjective assessment, occurs 6-7 times more often.

How to solve a triple integral?

The answer logically follows from the previous paragraph. Need to determine body traversal order and go to iterated integrals. Then deal with three single integrals sequentially.

As you can see, the whole kitchen is very, very reminiscent double integrals, with the difference that now we have added an additional dimension (roughly speaking, height). And, probably, many of you have already guessed how triple integrals are solved.

Let's dispel any remaining doubts:

Example 1

Please write down in a column on paper:

And answer the following questions. Do you know which surfaces define these equations? Do you understand the informal meaning of these equations? Can you imagine how these surfaces are located in space?

If you are inclined to the general answer “rather no than yes,” then be sure to work through the lesson, otherwise you will not advance further!

Solution: we use the formula.

In order to find out body traversal order and go to iterated integrals you need (everything ingenious is simple) to understand what kind of body this is. And in many cases, drawings greatly contribute to such understanding.

By condition, the body is limited by several surfaces. Where to start building? I suggest the following procedure:

First let's depict parallel orthogonal projection of the body onto the coordinate plane. First time I said what this projection is called, lol =)

Since projection is carried out along the axis, then first of all it is advisable to deal with surfaces, which are parallel to this axis. Let me remind you that the equations of such surfaces do not contain the letter "z". There are three of them in the problem under consideration:

– the equation specifies a coordinate plane that passes through the axis;
– the equation specifies a coordinate plane that passes through the axis;
– the equation sets plane "flat" straight line parallel to the axis.

Most likely, the desired projection is the following triangle:

Perhaps not everyone fully understood what we were talking about. Imagine that an axis comes out of the monitor screen and sticks directly into the bridge of your nose ( those. it turns out that you are looking at a 3-dimensional drawing from above). The spatial body under study is located in an endless trihedral “corridor” and its projection onto a plane most likely represents a shaded triangle.

I would like to draw special attention to the fact that while we have expressed just an assumption of projection and the clauses “most likely” and “most likely” were not accidental. The fact is that not all surfaces have been analyzed yet and it may happen that one of them “chops off” part of the triangle. As clear example suggests itself sphere with a center at the origin of radius less than one, for example, a sphere – its projection onto the plane (circle ) will not completely “cover” the shaded area, and the final projection of the body will not be a triangle at all (the circle will “cut off” its sharp corners).

At the second stage, we find out how the body is limited from above and from below and carry out a spatial drawing. Let's return to the problem statement and see which surfaces remain. The equation specifies the coordinate plane itself, and the equation – parabolic cylinder, located above plane and passing through the axis. Thus, the projection of the body is truly a triangle.

By the way, I found it here redundancy conditions - it was not necessary to include the equation of the plane, since the surface, touching the abscissa axis, already closes the body. It is interesting to note that in this case we would not be able to immediately draw the projection - the triangle would “draw” only after analyzing the equation.

Let's carefully depict a fragment of a parabolic cylinder:

After completing the drawings with the order of walking around the body no problem!

First, we determine the order of traversal of the projection (at the same time, it is MUCH MORE CONVENIENT to navigate using a two-dimensional drawing). It's done EXACTLY THE SAME, As in double integrals! Think of a laser pointer and scanning a flat area. Let's choose the “traditional” 1st bypass method:

Next, we pick up a magic lantern, look at the three-dimensional drawing and strictly from bottom to top We illuminate the patient. Rays enter the body through a plane and exit through the surface. Thus, the order of traversing the body is:

Let's move on to repeated integrals:

1) You should start with the “zeta” integral. We use Newton-Leibniz formula:

Let's substitute the result into the “game” integral:

What happened? Essentially, the solution was reduced to a double integral, and precisely to the formula volume of cylindrical beam! What follows is familiar:

2)

Pay attention to the rational technique for solving the 3rd integral.

Answer:

Calculations can always be written in “one line”:


But be careful with this method - gain in speed is fraught with loss of quality, and the more complex the example, the greater the chance of making a mistake.

Let's answer an important question:

Is it necessary to make drawings if the task conditions do not require their implementation?

You can go in four ways:

1) Draw the projection and the body itself. This is the most advantageous option - if you have the opportunity to complete two decent drawings, do not be lazy, do both drawings. I recommend it first.

2) Draw only the body. Suitable when the body has a simple and obvious projection. So, for example, in the disassembled example, a three-dimensional drawing would be enough. However, there is also a minus - it is inconvenient to determine the order of traversing the projection from a 3D picture, and I would recommend this method only to people with a good level of training.

3) Draw only the projection. This is also not bad, but then additional written comments are required, which limits the area from various sides. Unfortunately, the third option is often forced - when the body is too large or its construction is fraught with other difficulties. And we will also consider such examples.

4) Do without drawings at all. In this case, you need to imagine the body mentally and comment on its shape/location in writing. Suitable for very simple bodies or tasks where performing both drawings is difficult. But it’s still better to make at least a schematic drawing, since a “naked” solution may be rejected.

The following body is for independent work:

Example 2

Using a triple integral, calculate the volume of a body bounded by surfaces

In this case, the domain of integration is specified primarily by inequalities, and this is even better - a set of inequalities defines the 1st octant, including coordinate planes, and the inequality – half-space, containing the origin (check)+ the plane itself. The “vertical” plane cuts the paraboloid along the parabola, and it is advisable to construct this section in the drawing. To do this, you need to find an additional reference point, the easiest way is the vertex of the parabola (we consider the values and calculate the corresponding “zet”).

Let's continue to warm up:

Example 3

Using a triple integral, calculate the volume of the body bounded by the indicated surfaces. Execute the drawing.

Solution: The wording “execute a drawing” gives us some freedom, but most likely implies the execution of a spatial drawing. However, projection won’t hurt either, especially since it’s not the simplest here.

We stick to the previously proven tactics - first we’ll deal with surfaces, which are parallel to the applicate axis. The equations of such surfaces do not explicitly contain the variable “z”:

– the equation specifies the coordinate plane passing through the axis ( which on the plane is determined by the “eponymous” equation);
– the equation sets plane, passing through the “eponymous” "flat" straight line parallel to the axis.

The desired body is limited by a plane below and parabolic cylinder above:

Let’s create an order of traversal of the body, while the “X” and “Y” limits of integration, I remind you, it is more convenient to find out using a two-dimensional drawing:

Thus:

1)

When integrating over “y”, “x” is considered a constant, so it is advisable to immediately take the constant out of the integral sign.

3)

Answer:

Yes, I almost forgot, in most cases it is of little use (and even harmful) to check the result obtained with a three-dimensional drawing, since with a high probability volume illusion, which I talked about in class Volume of a body of revolution. So, evaluating the body of the problem considered, it seemed to me personally that it had much more than 4 “cubes”.

The following example for an independent solution:

Example 4

Using a triple integral, calculate the volume of the body bounded by the indicated surfaces. Make drawings of this body and its projection onto a plane.

An approximate example of a task at the end of the lesson.

It is not uncommon when the execution of a three-dimensional drawing is difficult:

Example 5

Using a triple integral, find the volume of a body given by its bounding surfaces

Solution: the projection here is not complicated, but you need to think about the order of traversing it. If you choose the 1st method, then the figure will have to be divided into 2 parts, which seriously threatens to calculate the sum two triple integrals. In this regard, the 2nd path looks much more promising. Let us express and depict the projection of this body in the drawing:

I apologize for the quality of some of the pictures, I cut them directly from my own manuscripts.

We choose a more advantageous order of traversing the figure:

Now it's up to the body. From below it is limited by the plane, from above - by the plane that passes through the ordinate axis. And everything would be fine, but the last plane is too steep and constructing the area is not so easy. The choice here is unenviable: either jewelry work on a small scale (since the body is quite thin), or a drawing about 20 centimeters high (and even then, if it fits).

But there is a third, native Russian method of solving the problem - to score =) And instead of a three-dimensional drawing, make do with a verbal description: “This body is limited by cylinders and a plane from the side, a plane from below and a plane from above.”

The "vertical" limits of integration are obviously:

Let's calculate the volume of the body, not forgetting that we bypassed the projection in a less common way:

1)

Answer:

As you noticed, bodies proposed in problems that are no more expensive than a hundred bucks are often limited by the plane below. But this is not a rule, so you always need to be on guard - you may come across a task where the body is located and under flat So, for example, if in the analyzed problem we instead consider the plane, then the examined body will be symmetrically mapped into the lower half-space and will be limited by the plane from below, and by the plane from above!

It's easy to see that you get the same result:

(remember that the body needs to be walked around strictly from bottom to top!)

In addition, the “favorite” plane may not be used at all, simplest example: a ball located above the plane - when calculating its volume, an equation will not be needed at all.

We will consider all these cases, but for now there is a similar task for you to solve on your own:

Example 6

Using the triple integral, find the volume of a body bounded by surfaces

A short solution and answer at the end of the lesson.

Let's move on to the second paragraph with equally popular materials:

Triple integral in cylindrical coordinates

Cylindrical coordinates are, in essence, polar coordinates in space.
In a cylindrical coordinate system, the position of a point in space is determined by the polar coordinates of the point - the projection of the point onto the plane and the applicate of the point itself.

The transition from a three-dimensional Cartesian system to a cylindrical coordinate system is carried out according to the following formulas:

In relation to our topic, the transformation looks like this:

And, accordingly, in the simplified case that we are considering in this article:

The main thing is not to forget about the additional “er” multiplier and place it correctly polar limits of integration when traversing the projection:

Example 7

Solution: we adhere to the same procedure: first of all, we consider equations in which the “ze” variable is absent. There is only one here. Projection cylindrical surface onto the plane represents the “eponymous” circle .

Planes they limit the desired body from below and above (“cut” it out of the cylinder) and project it into a circle:

Next up is a three-dimensional drawing. The main difficulty lies in constructing a plane that intersects the cylinder at an “oblique” angle, resulting in ellipse. Let us clarify this section analytically: to do this, we rewrite the equation of the plane in functional form and calculate the values ​​of the function (“height”) at the obvious points that lie on the boundary of the projection:

We mark the found points on the drawing and carefully (not like me =)) connect them with a line:

The projection of a body onto a plane is a circle, and this is a strong argument in favor of moving to a cylindrical coordinate system:

Let's find the equations of surfaces in cylindrical coordinates:

Now you need to figure out the order of traversing the body.

First, let's deal with projection. How to determine its traversal order? EXACTLY THE SAME AS WITH calculating double integrals in polar coordinates. Here it is elementary:

The “vertical” limits of integration are also obvious - we enter the body through the plane and exit it through the plane:

Let's move on to repeated integrals:

In this case, we immediately put the factor “er” into “our” integral.

As usual, a broom is easier to break along the twigs:

1)

We put the result into the following integral:

And here we do not forget that “phi” is considered a constant. But this is for the time being:

Answer:

A similar task for you to solve on your own:

Example 8

Calculate the volume of a body bounded by surfaces using a triple integral. Draw drawings of this body and its projection onto a plane.

An approximate sample of the final design at the end of the lesson.

Please note that in the conditions of the problems not a word is said about the transition to a cylindrical coordinate system, and an ignorant person will struggle with difficult integrals in Cartesian coordinates. ...Or maybe it won’t - after all, there is a third, original Russian way of solving problems =)

It's only begining! ...in a good way: =)

Example 9

Using the triple integral, find the volume of a body bounded by surfaces

Modest and tasteful.

Solution: this body is limited conical surface And elliptical paraboloid. Readers who have carefully read the article materials Basic surfaces of space, have already imagined what the body looks like, but in practice there are often more complex cases, so I will carry out a detailed analytical reasoning.

First, we find the lines along which the surfaces intersect. Let's compose and solve the following system:

From the 1st equation we subtract the second term by term:

The result is two roots:

Let's substitute the found value into any equation of the system:
, from which it follows that
Thus, the root corresponds to a single point - the origin. Naturally, because the vertices of the surfaces under consideration coincide.

Now let’s substitute the second root – also into any equation of the system:

What is the geometric meaning of the result obtained? “At height” (in the plane) the paraboloid and the cone intersect along circle– unit radius with center at point .

In this case, the “bowl” of the paraboloid contains the “funnel” of the cone, therefore forming The conical surface should be drawn with a dotted line (except for the segment of the generatrix farthest from us, which is visible from this angle):

The projection of a body onto a plane is circle with a center at the origin of radius 1, which I didn’t even bother to depict due to the obviousness of this fact (however, we do provide a written comment!). By the way, in the two previous problems, the projection drawing could also be scored, if not for the condition.

When moving to cylindrical coordinates using standard formulas, the inequality is written in its simplest form and there are no problems with the order of traversing the projection:

Let's find the equations of surfaces in a cylindrical coordinate system:

Since the problem considers the upper part of the cone, we express from the equation:

“We scan the body” from bottom to top. Rays of light enter it through elliptical paraboloid and exit through the conical surface. Thus, the “vertical” order of traversing the body is:

The rest is a matter of technique:

Answer:

It is not uncommon for a body to be defined not by its limiting surfaces, but by many inequalities:

Example 10

I explained the geometric meaning of spatial inequalities in sufficient detail in the same reference article - Basic surfaces of space and their construction.

Although this task contains a parameter, it allows for the execution of an exact drawing that reflects the basic appearance of the body. Think about how to build. A short solution and answer is at the end of the lesson.

...well, a couple more tasks? I was thinking about finishing the lesson, but I just feel like you want more =)

Example 11

Using a triple integral, calculate the volume of a given body:
, where is an arbitrary positive number.

Solution: inequality defines a ball with center at the origin of radius , and the inequality – the “inside” of a circular cylinder with an axis of symmetry of radius . Thus, the desired body is limited by a circular cylinder on the side and spherical segments symmetrical relative to the plane at the top and bottom.

Taking this as the base unit of measurement, let’s draw:

More precisely, it should be called a drawing, since I did not maintain the proportions along the axis very well. However, to be fair, the condition did not require drawing anything at all, and such an illustration turned out to be quite sufficient.

Please note that here it is not necessary to find out the height at which the cylinder cuts out the “caps” from the ball - if you take a compass in your hands and use it to mark a circle with a center at the origin of radius 2 cm, then the points of intersection with the cylinder will appear by themselves.

Transformation of double integral of rectangular coordinates, to polar coordinates
, related to rectangular coordinates by the relations
,
, is carried out according to the formula

If the domain of integration
limited to two beams
,
(
), coming out of the pole, and two curves
And
, then the double integral is calculated using the formula

.

Example 1.3. Calculate the area of ​​the figure bounded by these lines:
,
,
,
.

Solution. To calculate the area of ​​an area
Let's use the formula:
.

.

1.2. Triple integrals

The basic properties of triple integrals are similar to the properties of double integrals.

In Cartesian coordinates, the triple integral is usually written as follows:

.

If
, then the triple integral over the area numerically equal to the volume of the body :

.

Triple integral calculation

Let the domain of integration bounded below and above, respectively, by single-valued continuous surfaces
,
, and the projection of the region to the coordinate plane
there is a flat area
(Fig. 1.6).

.

Example 1.4. Calculate
, Where - a body limited by planes:

Triple integral in cylindrical coordinates

When moving from Cartesian coordinates
to cylindrical coordinates
(Fig. 1.9) associated with
relations
,
,
, and

Region
, bounded by a curve
, will take the form, or
, while the polar angle
. As a result we have

.

2. Elements of field theory

Let us first recall the methods for calculating curvilinear and surface integrals.

Calculation of a curvilinear integral over coordinates of functions defined on a curve , reduces to calculating a definite integral of the form

if the curve specified parametrically
corresponds to the starting point of the curve , A
- its end point.

Calculation of the surface integral of a function
, defined on a two-sided surface , comes down to calculating a double integral, for example, of the form

if the surface , given by the equation
, is uniquely projected onto the plane
to the region
. Here - angle between unit normal vector to the surface and axis
:

Side of the surface required by the problem conditions is determined by the choice of the appropriate sign in formula (2.3).

Definition 2.1. Vector field
called the vector function of a point
along with its scope:

Vector field
characterized by a scalar quantity – divergence:

Definition 2.2. Flow vector field
through the surface called the surface integral:

Where - unit normal vector to the selected side of the surface , A
- scalar product of vectors And .

Definition 2.3. Circulation vector field

By closed curve called a curvilinear integral

Where
.

Ostrogradsky-Gauss formula establishes a connection between the vector field flow through a closed surface and field divergence:

Where - surface bounded by a closed contour , A is the unit normal vector to this surface. The direction of the normal must be consistent with the direction of the contour traversal .

Example 2.1. Calculate Surface Integral

,

Where - outer part of the cone
(
), cut off by a plane
(Figure 2.1).

Solution. Surface uniquely projected into the region
plane
, and the integral is calculated using formula (2.2).

Region
there is a circle
. Therefore, in the last integral we move to polar coordinates, while
,
:

Example 2.2. Find the divergence and curl of a vector field
.

Solution. Using formula (2.4) we obtain

The rotor of a given vector field is found using formula (2.5)

Example 2.3. Find the flux of a vector field
through part of the plane :
, located in the first octant (the normal forms sharp corner with axle
).

Where
- plane projection on
(Fig. 2.4).

Example 2.4. Compute the flux of a vector field through a closed surface , formed by the plane
and part of the cone
(
) (Fig. 2.2).

Solution. Let us use the Ostrogradsky-Gauss formula (2.8)

.

Let's find the divergence of the vector field according to formula (2.4):

Where
is the volume of the cone over which integration is carried out. Let's use the well-known formula to calculate the volume of a cone
(- radius of the base of the cone, - his high). In our case we get
. Finally we get

.

Example 2.5. Calculate the circulation of a vector field
along the contour , formed by the intersection of surfaces
And
(
). Check the result using the Stokes formula.

Solution. The intersection of these surfaces is a circle
,
(Fig. 2.1). The traversal direction is usually chosen so that the area limited by it remains to the left. Let's write the parametric equations of the contour :

and the parameter varies from before
. Using formula (2.7), taking into account (2.1) and (2.10), we obtain

.

Let us now apply the Stokes formula (2.9). As a surface , stretched on the contour , you can take part of the plane
. Normal direction
to this surface is consistent with the direction of the contour traversal . The curl of a given vector field is calculated in Example 2.2:
. Therefore, the desired circulation

Where
- area of ​​the region
.
- circle radius
, where

Examples of solutions of arbitrary triple integrals.
Physical applications of the triple integral

In the 2nd part of the lesson we will work out the technique of solving arbitrary triple integrals , whose integrand function of three variables in the general case it is different from a constant and continuous in the region ; and also get acquainted with the physical applications of the triple integral

I recommend that new visitors start with part 1, where we covered the basic concepts and the problem of finding the volume of a body using a triple integral. I suggest the rest of you repeat it a little. derivatives of functions of three variables, since in the examples of this article we will use the inverse operation - partial integration functions

In addition, there is one more important point: if you are not feeling well, then it is better to postpone reading this page if possible. And the point is not only that the complexity of calculations will now increase - most triple integrals do not have reliable ways manual checks, so it is highly undesirable to start solving them in a tired state. For low tone it is advisable solve something easier or just relax (I’m patient, I’ll wait =)), so that another time with a fresh head I can continue to crack down on triple integrals:

Example 13

Calculate triple integral

In practice, the body is also denoted by the letter , but this is not very a good option, in view of this, “ve” is “reserved” for the designation of volume.

I’ll tell you right away what NOT to do. No need to use linearity properties and represent the integral in the form . Although if you really want to, then you can. In the end, there is a small plus - although the recording will be long, it will be less cluttered. But this approach is still not standard.

In the algorithm solutions there will be little novelty. First you need to understand the domain of integration. The projection of the body onto a plane is a painfully familiar triangle:

The body is limited from above plane, which passes through the origin. By the way, you need to first be sure to check(mentally or in draft), whether this plane “cuts off” part of the triangle. To do this, we find its line of intersection with the coordinate plane, i.e. we decide the simplest system: - no, this one straight (not on the drawing)“passes by”, and the projection of the body onto the plane really represents a triangle.

The spatial drawing here is not complicated either:

In fact, it was possible to limit it to only this, since the projection is very simple. ...Well, or just a projection drawing, since the body is also simple =) However, not drawing anything at all, I remind you, is a bad choice.

Well, of course, I can’t help but please you with the final task:

Example 19

Find the center of gravity of a homogeneous body bounded by surfaces, . Draw drawings of this body and its projection onto a plane.

Solution: the desired body is limited by the coordinate planes and the plane, which is convenient for subsequent construction present in segments: . Let’s choose “a” as the scale unit and make a three-dimensional drawing:

The drawing already has a ready-made center of gravity point, however, we don’t know it yet.

The projection of a body onto a plane is obvious, but, nevertheless, let me remind you how to find it analytically - after all, such simple cases do not always occur. To find the line along which the planes intersect, you need to solve the system:

Substitute the value into the 1st equation: and we get the equation "flat" straight:

We calculate the coordinates of the center of gravity of the body using the formulas
, where is the volume of the body.