The procedure for calculating the lossy integral is similar to the general operation of the running integral. For її description, we introduce the understanding of the correct trivial area:

Appointment 9.1. Trivial region V, surrounded by a closed surface S, is called regular, because:

1. be straight, parallel to axis Oz that is drawn through the internal point of the region, crossing S at two points;
2. the entire region V is projected onto the Oxy plane in a regular two-world region D;
3. whether a part of area V, visible in it by a plane, parallel to whether it is from the coordinate planes, may have power 1) and 2).

Let's look at the correct area V, I will border the bottom and the top with surfaces z=χ(x,y) and z=ψ(x,y) and projected onto the Oxu y plane, the correct area D, the middle of which x will change in the boundaries from a to b, will be surrounded by curves y=φ1(x) and y=φ2(x) (Fig. 1). Let f(x, y, z) be a continuous function in the domain V.

Appointment 9.2. It is called the three-fold integral of the function f(x, y, z) over the region V in the form:

Trirazovy іntegrа maє tі zh vlastivostі, shcho і dvorazovy. Pererakhuyemo їх without confirmation, shards of stench are brought up similarly to the fall of the yard-wise integral.

Calculation of the lossy integral.

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

. (9.3)

Bringing.

Rozіb'ёmo area V planes, parallel to the coordinate planes, on n regular areas. Todі z power 1 yelling

where is the three-time integral of the function f(x,y,z) in the domain .

Vikoristovuyuchi formula (9.2), forward parity can be rewritten at a glance:

Understanding the continuity of the function f (x, y, z) is clear, which is the boundary of the integral sum, which stands at the right side of the equation of equality, and is equal to the third integral. Then, passing to the boundary when, we take:

what it was necessary to bring.

Respect.

In a similar way to the fall of the undercurrent integral, one can bring that changing the order of integration does not change the value of the three-time integral.

butt. Calculating the integral de V is a triangular pyramid with vertices at the points (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1). Її projection onto the Oxy plane є tricutnik with vertices (0, 0), (1, 0) and (0, 1). From the bottom, the area is bordered by the area z = 0, and from the top – by the area x + y + z = 1. Let's move on to the triple integral:

Multipliers, which do not lie in changeable integration, can be blamed for the sign of the double integral:

Curvilinear coordinate systems in the trivial space.

1. Cylindrical coordinate system.

Cylindrical coordinates of the point Р(ρ,φ,z) – cepolar coordinates ρ, φ of the projection of the point on the Ohu plane and the applicator of the given point z (Fig. 2).

The formulas for the transition from cylindrical coordinates to Cartesian coordinates can be set as follows:

x = ρ cosφ, y = ρ sinφ, z = z. (9.4)

1. Spherical coordinate system.

For spherical coordinates, the position of the point in space is indicated by the linear coordinate ρ - the distance from the point to the cob of the Cartesian coordinate system (or the poles of the spherical system), φ - the polar edge between the positive pіvvіssyu Ox and the projection of the point onto the Oxy plane, and θ - the kutom between the positive Oz and double OP (Fig. 3). With whom

Given the formula for the transition from spherical coordinates to Cartesian:

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

Jacobian and yogo geometrical zmist.

Let's look at the wild trend of replacing the changes in the subway integral. Nehai at the Ohu flat area D is given, surrounded by a line L. Assume that х і у є are single-valued and continuously differentiating functions of new changing u and v:

x = φ(u, v), y = ψ(u, v). (9.6)

Let's look at the rectangular coordinate system Ouv, the point P(u, v) which points P(x, y) from the region D. All such points form the region D near the plane Ouv, I'm surrounded by a line L?. It can be said that formulas (9.6) establish a one-to-one correspondence between points of the regions D and D. For which lines u = const that

v = const at the Ouv plane will be similar to the lines at the Ohu plane.

We can see in the Ouv plane a rectangular maidan ΔS, bordered by straight lines u = const, u + Δu = const, v = const і v + Δv = const. Їy vіdpovidatimé curvilinear maidanchik ΔS near the Ohu flat (Fig. 4). The areas of analysis of Maidanchiks will be denoted as ΔS and ΔS. For ciomu ΔS = Δu Δv. We know the area ΔS. Significantly, the vertices of the curvilinear chotyrikutnik P1, P2, P3, P4 de

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).

Replacing small zbіlshennya Δu і Δv vіdpovіdmi differentials. Todi

With which chotirikutnik P1 P2 P3 P4 can be taken as a parallelogram and the area can be assigned to the formula for analytical geometry:

(9.7)

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

Passing to the border with equality (9.7), we take away the geometric Jacobian shift:

so the Jacobian module is the boundary between the area of ​​infinitely small squares S and S.

Respect. By a similar rank, one can designate the understanding of the Jacobian and its geometrical meaning for the n-world space: that x1 = φ1(u1, u2,…,un), x2 = φ2(u1, u2,…,un),…, xn = φ(u1 , u2, ..., un), then

(9.8)

With this, the Jacobian modulus gives a boundary between "obsyagiv" small areas of space x1, x2, ..., xn and u1, u2, ..., un.

Replacing Changes in Multiple Integrals.

Dolіdzhuєmo zagalny vpadok zameni zmini z butt podvіynogo іntegral.

Let a continuous function z = f(x,y) be given in the area D, the same value of the function z = F(u, v) in the area D, de

F(u, v) = f(φ(u, v), ψ(u, v)). (9.9)

Let's look at the integral sum

The deintegral sum on the right is taken over the region D. Passing to the boundary when we take away the formula for transforming the coordinates in the sweeping integral.

Try the integrals. Calculation of the volume of the body.
Trial integral in cylindrical coordinates

Three days in the dean's office the sky lay, at the pants of Pіthagoras robes,
In the hands of Fikhtengolts, there is a volume of trimavs, that the yogi of the white light is alive,
To nіg they tied the third integral, and wrapped the corpse in the matrix,
And the deputy of prayer is like a nahabnik after reading Bernoulli's theorem.

Lost integrations are those that you can’t be afraid of anymore =) Because if you read the whole text, then it’s better for everything you’ve got it wrong theory and practice of "superior" integrals, as well as dependent integrals. And there, de podvіyny, nearby and lost:

Indeed, what is there to be afraid of? The integral is smaller, the integral is larger.

Let's take a look at the record:

- the icon of the trinity integral;
- Pidіntegralna triple change function;
- Dobutok differentials.
- Area of ​​integration.

Particularly noteworthy for gallery integration. Yakscho in underlined integral won flat figure, then here - expanse body , yaka, ya know on top. In this rank, the crime of the viscerally guessed you is guilty of orienting yourself in main surfaces and remember to win the simplest trivimir armchairs.

The deyakians were embarrassed, wise…. Unfortunately, the article cannot be called “consumed integrals for dummies”, and it is necessary to know / remember something. Ale, nothing terrible - all the material of publications in a borderline accessible form will be mastered in the shortest term!

What does it mean to calculate the lost integral, and what did it take?

Calculate the lost integral - tse means know KILO:

In the simplest way, if, the third integral is numerically more advanced in relation to the body. І deisno, vіdpovіdno to integration, tvir one infinitely small the volume of the elementary "ceglinka" of the body. And the third integral is united all qi infinitely small particles by region, after which the integral (total) value of the volume of the body comes out: .

In addition, the third integral is important physical programs. Ale about tse pіznіshe - in the 2nd part of the lesson, dedication calculation of additional losses of integrals, for which the function of the variable is constant as a constant and is uninterrupted in the sphere. In this article, we can see in detail the meaning of the obyagu, as my subjective assessment is seen in 6-7 times more often.

How to solve the lost integral?

Vіdpovіd is logically viplivає from the previous paragraph. It is necessary to appoint body bypass order i go to we repeat the integrals. After that, successively solve with three single integrals.

Yak bachite, the whole kitchen is more and more nagaduє underlying integrals, From the tієyu vіdminnіstyu, scho at the same time we have got additional rozmirnіst (roughly seeming, height). I, singly, many of you have already guessed how the losses of the integrals are violating.

Let's sum up what we have lost:

butt 1

Be kind, rewrite with a stamp on paper:

І give advice on the next meal. Chi know You, what are the surfaces to set qi equal? Chi zrozumіly you informal zmіst tsikh rivnyan? Chi yavlyaєєєєєєєєєєєV, yak і surface raztashovanі in space?

If you shilyayetsya to zagalnoї vіdpovіdі "more nі, nizh so", then obov'yazkovo opratsyut lesson, otherwise you won't get farther!

Solution: vicorist formula

In order to schob z'yasuwati body bypass order i go to we repeat the integrals it is necessary (everything is ingeniously simple) to understand what it was. And it’s great to put armchairs on such a rose in rich vipadkas.

Behind the mind, the body is surrounded by kilkom surfaces. Why start chic? I pronounce the next order diy:

On the cob is imaginable parallel orthogonal projection of the body onto the coordinate plane. The first time I said, what is the name of the projection, lol =)

If the design is to be carried out on a grand scale, then in Persh surfaces, yakі parallel to tsієї axis. I guess what kind of surfaces such do not revenge the letters "ze". The examined manager has three:

- Rivnyannya sets the coordinate area, how to pass through the whole;
- Rivnyannya sets the coordinate area, how to pass through the whole;
- equal task flat "flat" straight parallel to the axis.

Shvidshe for everything, shukana projection є coming trikutnik:

Possibly, not everyone had a residual understanding of where to go. Show that everything comes out of the monitor screen and sticks right in your transfer ( tobto. come out, you marvel at the 3rd world chair of the beast). Doslіdzhuvane expanses of the body are found in the non-skinned trihedral "corridor" and its projection onto the area of ​​the naimovіrnіshe є shaded tricutnik.

I give special respect to what we hung out more excuse about projection and the warning “neishvidshe”, “nayimovirnishe” were vipadkovy. On the right, in that not all surfaces have been analyzed yet, and it can be so, that even from them “discover” a part of the tricutnik. Like a primer butt you ask sphere centered on the cob of coordinates with a radius less than one, for example, a sphere – її projection on the plane (column ) I won’t repeat the “nakry” shaded area, and the projection of the body will be called not a tricot (kolo "zrіzhe" youmu gostrі kuti).

From the other side of the stage, it’s z’yasovuєmo, chim the body is surrounded by the beast, lower from the bottom and vikonuemo the expanse of the armchair. We turn to the mind and marvel at the surface, as if the surface was gone. Leveling sets the coordinate plane itself, and leveling - parabolic cylinder, retasting above flat and pass through the whole. In this rank, the projection of the body is diisno є trikutnik.

Before speech, here appeared supermundane think - in the new bulb, it’s not obov’yazkovo to include even planes, shards of the surface, sticking out the abscissa axis, and so the body closes. It means that in this particular moment we would not have been able to christen the projection - the tricutnik “drawn” only after the analysis of equalization.

A fragment of a parabolic cylinder is accurately depicted:

After the vikonannya armchair z bypassing the body no problems!

On the back of the head, it is significant the order in which the projection is traversed (with the help of the best hand, be guided by two-world armchairs). Tse shy ABSOLUTELY SO, yak i in lower integrals! Guessing laser pointer that scanning of a flat area. Choose "traditional" 1st bypass method:

Dali take in the hands of the charming lighter, marveling at the trivimir of the armchair and strictly downhill enlighten the patient. Changes to enter the body through the surface and exit from it through the surface. In this order, the order of bypassing the body:

Let's move on to repeated integrations:

1) Start the following from the "Z" integral. Vikoristovuemo Newton-Leibniz formula:

Imagine the result of the "igame" integral:

What happened? As a matter of fact, the solution was reduced to a sub-integral, and itself - to a formula. volume of cylindrical beam! More well know:

2)

Give respect to the rational technique of solving the 3rd integral.

Vidpovid:

The calculation can be written down and “in one row”:


But in this way, be careful - if you win at swidkost, you threaten with a waste of time, and if you have an important butt, there are more chances for a pardon.

Note on important nutrition:

Why is it necessary to work an armchair, so that the head of the mind does not require their vikonannia?

You can drink chotirma with paths:

1) Draw the projection of the same body. The best option is that it is possible to vikonate two decent armchairs, do not lament, rob offended armchairs. I recommend us forward.

2) Draw more body. Suitable, if the body is clumsy, that obvious projection. So, for example, a triple-breasted armchair was stuck at the selected butt. However, here is the minus - according to the 3D-picture, it is not handy to determine the order of bypassing the projection, and this way I am only happy for people with a good level of training.

3) Show more projection. Tezh not bad, but about obov'yazkovі dodatkovі pisletovі komentari, nizh zamezhena region from raznih storіn. Unfortunately, the third option is often confusing - if it's too late, it's too big to deal with other difficulties. І takі apply mi so razglyademom.

4) Get around without an armchair. It is necessary for each person to present the body of the thought and to comment on the form / format in writing. It’s good to go for the simplest ones tіl chi zavdan, de vikonannya both the armchair is important. But all the same, it’s better if you want to use sketchy little ones, shards of a “goal” solution can be rejected.

Come body for independent help:

butt 2

For the help of the loss integral, calculate the volume of the body, surrounded by surfaces

At to this particular type the area of ​​integration is given more importantly by irregularities, and the price is more short - without any irregularities sets the 1st octant, including the coordinate planes, and the unevenness - napіvspіr, how to avenge the cob of coordinates (reverse)+ the area itself. The “vertical” plane is spread out by a paraboloid parabola and on the armchair bazhan it is necessary to induce dandelions. For whom it is necessary to know the additional reference point, in simpler terms, the top of the parabola. (We can see the meaning and rozrakhovuyemo vіdpovіdne "z").

Let's continue to understand:

butt 3

Calculate for the help of the loss integral the volume of the body surrounded by designated surfaces. Vikonati armchair.

Solution: the formula "viconati of the armchair" gives us deak freedom, ale, better for everything, transferring the vikonanny of a spacious armchair. However, the projection can’t be wound up either, it’s not the easiest thing to do here.

Dotrimuёmosya vіdpratsovanoї earlier tactics surfaces, as if parallel to the axis of the application. The equalization of such surfaces should not be avenged by clearly changing the “Z”:

- Rivnyannya sets the coordinate plane to pass through the whole ( yak on the flat is assigned to the "same name" equals);
- equal task flat, to pass through the "same line" "flat" straight parallel to the axis.

The body that is joking is surrounded by a flat bottom and parabolic cylinder beast:

Let's put together a procedure for bypassing the body, with which “iksovі” and “igrokovі” interintegration, I guess, it’s better to sing behind the two-world armchairs:

In this manner:

1)

When integrating behind the "iplayer" - "ix" is considered a constant, then the constant should be blamed for the sign of the integral.

3)

Vidpovid:

So, without forgetting a little, zdebіlshogo otmany the result of little (and navit shkіdlivo) zvіryati z trivimirnym armchairs, oskolki z great ymovіrnіstyu vinikne illusion oblige, About yaku I rozpov_shche at the lesson Volume body wrap. So, estimating the body of the looked-out leader, I was especially lucky that there are more than 4 “cubes” in the new one.

An offensive butt for an independent vision:

butt 4

Calculate for the help of the loss integral the volume of the body surrounded by designated surfaces. The work of the armchair of this body and its projection on the plane.

Zrazok designed as a task for a lesson.

It is not rare, if the vikonnanny of the trivimir chair is more difficult:

butt 5

For the help of the lossy integral, to know the volume of the body, given by the surfaces, which surround it.

Solution: the projection here is clumsy, but over the order of the bypass, you need to think How to choose the 1st method, then the figure will have to be divided into 2 parts, which will inevitably threaten the calculation of sumi two Trinity integrals. For someone with a richer perspective, there is another path. It can be seen and visualized by the projection of this body on the armchair:

I will ask again for the accuracy of such pictures, I wirl them directly from my own manuscripts.

We choose a more viable order for bypassing the figure:

Now to the right behind the body. From below, it is surrounded by a flat area, from the beast - by a flat area, so as to pass through the entire ordinate. And everything would be nothing, but the rest of the flat is too steep and it’s not so easy to get around the area. The choice here is unenviable: either the jewelry robot is on a small scale (because it was thin to make it thin), or the armchair is about 20 centimeters high (that and those that can fit).

Ale and the third, calmly Russian method of solving the problem is to score =) and a flat to the side, a flat to the bottom, and a flat to the beast.

"Vertical" inter-integration is obviously like this:

Let's calculate the volume of the body, not forgetting that we got around the projection in a smaller expanded way:

1)

Vidpovid:

As you remember, proponing in the zavdannya of the body is not expensive for a hundred bucks, often surrounded by a flat below. But it’s not a rule, so you need to be ready - you can spend the day, de tilo roztashovani pid flat. So, for example, if you look at the flat in the selected zamіst, then the body will be symmetrically represented in the lower space and will be surrounded by a flat from below, and by a flat to the beast!

It's easy to switch over to see the same result:

(Remember that it is necessary to get around strictly downhill!)

In addition, "in love" with the flat can appear in front not on the right, the simplest butt: a sack, stashed more than the flat - with the calculation of the yogic obligation, you do not need to look in front.

We can see all these views, but for the time being, the task for an independent vision is similar:

butt 6

For the help of the lossy integral, to know about the body, surrounded by surfaces

Briefly, the solution is to illustrate the lesson.

Let's move on to another paragraph with no less popular materials:

Trial integral in cylindrical coordinates

Cylindrical coordinates - ce, in fact, 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 applique of the point itself.

The transition from the trivi- mer Cartesian system to the cylindrical coordinate system is carried out by the following formulas:

One hundred and fifty of our transformations look like this:

I, apparently, in a simple way, which is easily seen in this article:

Golovne, do not forget about the additional multiplier "er" and correctly arrange polarity between integration when bypassing the projection:

butt 7

Solution: dotrimuєmosya of the same order diy: we look ahead to the equal, in some days the “Z” is changed There is only one here. projection cylindrical surface on the area є "of the same name" colo .

Squares surround the shukane body from the bottom and the beast (“hang” yoga from the cylinder) and are designed in the colo:

On a black trivimir armchair. The main difficulty lies in the surface area, as if the cylinder is twisted under the “slanted” hood, after which to go elips. Let's clarify this rewriting analytically: for which we rewrite the plane of the functional view and we calculate the value of the function (“height”) at the points that we ask, as if to lie on the interprojection:

Looks like you know the points on the armchair and carefully (and not like that, like me =)) zadnuєmo їх line:

The projection of the body onto the plane is the length, and the length of the argument for the speed of the transition to a cylindrical coordinate system:

We know the alignment of the surface at cylindrical coordinates:

Now follow the procedure for bypassing the body.

Let's take a look at the back of the head from the projection. How to determine the order of bypass? Exactly like that calculation of sub-integrals in polar coordinates. Here wine is elementary:

"Vertical" interintegration is also obvious - it enters the body through the plane and exits it through the plane:

Let's move on to repeated integrations:

For which multiplier "er" is immediately put in the "own" integral.

Vinik yak zavzhd is easier to break through the twigs:

1)

We take the result of the offensive integral:

And here it is not forgotten that fi is important as a constant. Ale tse until the singing hour:

Vidpovid:

Similar task for an independent vision:

butt 8

Calculate for the help of the loss integral the volume of the body surrounded by surfaces. Vikonati armchair of this body and its projection on the square.

Zrazok fine design like a lesson.

To be sure, that in the minds of the problems of the same word it is not said about the transition to a cylindrical coordinate system, and the person will not be known to struggle with important integrals in Cartesian coordinates. ... Or maybe it won’t be - even if it’s the third, calmly Russian way of solving problems.

All just start! …in a good sense: =)

butt 9

For the help of the lossy integral, to know about the body, surrounded by surfaces

Modestly and with relish.

Solution: whole finite surfaceі elliptic paraboloid. Readers, who are respectfully familiar with the materials of the article Main surfaces of space, already presented, as if looking at the body, but in practice folded vipads often trap, so I will conduct a report on the analytical world.

We know the lines on the back, with which the surfaces are tinted. We build and build the following system:

From the 1st equal, we can see each other in terms of terms:

As a result, two roots are taken away:

Imagine knowing the meaning of whether the system is equal:
the stars are screaming
Otzhe, root vіdpovіdaє single point - the cob of coordinates. Naturally - even the tops of the tops of the tops run up.

Now let's imagine another root - the same for whether the system is equal:

What is the geometrical replacement for the result? "On the heights" (near the plane) the paraboloid and the cone are tinted along cola- single radius centered at the point.

When the "cup" of the paraboloid contains the "funnel" of the cone, appease the final surface should be crossed with a dotted line (behind the vine it is a distant view of us, as seen from this angle):

The projection of the body on the plane colo with the center on the cob of coordinates of radius 1, which I did not dare to depict through the obviousness of this fact (prote letter comment robimo!). Before the speech, at the two front chairs on the chair, the projections could be beaten, yakby not mind.

When moving to cylindrical coordinates, the standard formulas can be used to write down the unevenness in the simplest way and in order to bypass the projection of everyday problems:

We know the surface alignment of the cylindrical coordinate system:

Since the problem looks at the upper part of the cone, then it can be seen:

"Scanuemo body" from the bottom uphill. Change the light to enter before the new through elliptic paraboloid and exit through the end surface. In this order, the "vertical" order of bypassing the body:

Second right technique:

Vidpovid:

It’s not rare, if the body is asked not to surround it with surfaces, but without any irregularities:

butt 10

Geometric zmist expanse of irregularities, I reportedly explained from the same proving article. The main surfaces of the space.

Tse zavdannya want and stash the parameter, but allow for the exact armchair, which inspires an important appearance of the body. Think like a vikonati pobudova. In short, the solution is to prove it - like a lesson.

... well, what, is it a sprat? Thinking about finishing the lesson, but then I guess what you want more =)

butt 11

For the help of the lossy integral, calculate the volume of the given body:
, De - More positive number.

Solution: unevenness set the column with the center on the cob of coordinates to the radius, and the unevenness - "Internality" of a circular cylinder with all the symmetry of the radius. In this order, the body, as if whispering, is surrounded by a circular cylinder on the side and spherical segments symmetrical to the surface at the top and bottom.

Taking for the basic unit of the world, we take the armchair:

More precisely, yoga should be called a little baby, the shards of proportion along the axis I won’t be any better. Prote, for the sake of justice, for the mind, it was not necessary to raise anything, and such an illustration appeared to be quite sufficient.

To show respect, that here it’s not obov’yazkovo z’yasovuvati height, on such a cylinder hanging from the back of the “hat” - just pick up a compass and mark a column with the center on the cob of coordinates with a radius of 2 cm, then the points of the crossbar with the cylinder will appear by themselves .

1. Cylindrical coordinates are a set of polar coordinates at the xy plane and from the sig- nificant Cartesian applicator z (Fig. 3).

Let M(x, y, z) be a sufficient point in the space xyz, P is the projection of the point M onto the plane xy. The point M is uniquely assigned by a trinity of numbers - the polar coordinates of the point P, z - the applicate of the point M. The formulas that call them Cartesian may look

Jacobian fermentation (8)

butt 2.

Calculate the integral

de T - area surrounded by surfaces

Solution. We pass in the integral to spherical coordinates using formulas (9). Same area of ​​integration can be set with irregularities

And that means

butt 3 Know the volume of the body, fringed:

x 2 + y 2 + z 2 \u003d 8,

Maemo: x 2 + y 2 + z 2 \u003d 8 - a sphere with a radius R \u003d v8 with a center at the point O (000),

The upper part of the cone z2 = x2 + y2 with all the symmetry of Oz and the apex at the point O (Fig. 2.20).

We know the line of the crossbar of the sphere of the cone:

І shards for the mind z? 0, then

Circle R=2, which lies near the plane z=2.

That's right (2.28)

de area u bordered by the beast

(part of a sphere),

(Part of a cone);

area U is projected on the Ohu area area D - radius 2.

Also, to pass incrementally in the integral to the cylindrical coordinates, victorious formulas (2.36):

Between changes, r is significant according to the distance D v outside the column R=2 with the center at the point O, by the same token: 0?c?2p, 0?r?2. In this way, the region U in cylindrical coordinates is marked by advancing irregularities:

We respect that

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Potential integral.

Control food.

Consistent integral, yoga of power.

Replacement of changes in the third integral. Calculation of the lossy integral in cylindrical coordinates.

Calculation of the lossy integral in spherical coordinates.

Come on function u= f(x,y,z) assigned to the closed area V space R 3 . Rozib'ёmo region V decent rank on n elementary closed areas V 1 , … ,V n, scho V 1 , …,  V n obviously. Significantly d- the largest of the diameters of the regions V 1 , … ,V n. At the skin area V k choose a good point P k (x k ,y k ,z k) and warehousing integral sum functions f(x, y,z)

S =

Appointment.Trial integral type of function f(x, y,z) by region V called the inter-integral sum
yakscho vin isnuє.

in such a manner,

(1)

Respect. Integral sum S deposit in the way of breaking up the region V ta select point P k (k=1, …, n). However, if there is a boundary, then it won’t lie in the way of breaking up the region V ta select point P k. If you compare the designation of the subvariant and the incremental integrals, then it is easy to use the same analogy in them.

Sufficient reasoning for the lossy integral. Trial integral (13) is used as a function f(x, y,z) is rimmed in V i is uninterrupted in V, behind the crown of the final number of lumpy-smooth surfaces, rotting at V.

Acts of power of the potry integral.

1) Yakscho W- Numerical constant, then

3) Additiveness by region. Yakscho region V divided into regions V 1 і V 2 , then

4) Fit body V dorivnyuє

(2 )

Calculation of the lossy integral in Cartesian coordinates.

Come on D body projection V on the flat xOy, surface z=φ 1 (x,y),z=φ 2 (x, y) surround the body V below that beast is clear. Tse means what

V = {(x, y, z): (x, y)D , φ 1 (x,y)≤ z ≤ φ 2 (x,y)}.

Such a body is called z- Cylindrical. Trial integral (1) z-cylindrical body V calculated by the transition to the repeated integral, which is added up from the hinged type of the integral:

(3 )

In this repeated integral of the backbone, the internal sing integral of the change is calculated z, at which x, y vvazhayutsya imminent. Let's count lower integral view of the selected function by region D.

Yakscho V x- cylindrical or y- cylindrical body, then correct to the formula

For the first formula D  body projection V to the coordinate plane yOz, and in the other - on the plane xOz

apply. 1) Calculate the body total V, surrounded by surfaces z = 0, x 2 + y 2 = 4, z = x 2 + y 2 .

Solution. Let's count for the help of the loss integral behind the formula (2)

Let's move on to the repeated integral after formula (3).

Come on D- colo x 2 +y 2 ≤ 4, φ 1 (x , y ) = 0, φ 2 (x , y )= x 2 +y 2. Todi following the formula (3) is taken

To calculate this integral, we pass to polar coordinates. When tsimu kolo D transform into faceless

D r = { (r , φ ) : 0 ≤ φ < 2 π , 0 ≤ r ≤ 2} .

2) Tilo V surrounded by surfaces z=y , z=-y , x= 0 , x= 2, y= 1. Calculate

Squares z=y , z=-y to encircle the body from below and to the beast, flat x= 0 , x= 2 encircle the body in the back and front, and the flat y= 1 right hand V-z- cylindrical body, yoga projection D on the flat hoyє rectangle OABC. Let's put it down φ 1 (x , y ) = -y

Reworking of the vertical integral in the form of rectangular coordinates to polar coordinates
, connected with right-angled coordinates
,
, follow the formula

What is the area of ​​integration
surrounded by two exchanges
,
(
), which go out of the poles, those two are crooked
і
, then the underlying integral is calculated using the formula

.

butt 1.3. Calculate the area of ​​​​the figure, surrounded by these lines:
,
,
,
.

Solution. For calculating the area of ​​the region
accelerated by the formula:
.

Imaginable area (Fig. 1.5). For whom we convert curves: , , , . Let's move on to polar coordinates: , . . In the polar coordinate system, the area be described by equals:

.

1.2. Potential integrals

The main powers of the third integrals are analogous to the powers of the lower integrals.

In Cartesian coordinates, the third integral should be written as follows:

.

Yakscho
, then the third integral over the region numerically greater volume of the body :

.

Calculation of the lossy integral

Let the area of ​​integration it is surrounded from below and to the beast by clearly unambiguous uninterrupted surfaces
,
, moreover, the projection of the area to the coordinate plane
є flat area
(Figure 1.6).

Same with fixed values valid applications area point change at the borders. Todi otrimuemo: . What, moreover, a projection signify irregularities , , de - unambiguous uninterrupted functions on the , then

.

butt 1.4. Calculate
, de - solid, surrounded by flats:

	, , , ( , , ). Solution. The area of ​​integration is the pyramid (Fig. 1.7). Area projection є trikutnik , by straight lines , , (Fig. 1.8). At Applique dot satisfy the nervousness to that .
	Arranging interintegration for tricoutnik , taken

Trial integral in cylindrical coordinates

When switching to Cartesian coordinates
to cylindrical coordinates
(Fig. 1.9), pov'yazanih z
spіvvіdneshennya
,
,
, moreover

	, ,, the third integral is converted to: butt 1.5. Calculate the volume of the body, surrounded by surfaces: , , . Solution. Volume of the body, what to joke about dorivnyuє .
	The areas of integration are part of a cylinder, surrounded by a flat bottom. , but the beast is flat (Figure 1.10). Area projection є kolo centered on the cob of coordinates and with a single radius. Let's move on to cylindrical coordinates. , , . At Applique dot , satisfies the nervousness or in cylindrical coordinates:

Region
, surrounded by a curve
, look forward to seeing, otherwise
at tsimu polar kut
. May the results

.

2. Elements of field theory

Let's guess ahead of time how to calculate curvilinear and surface integrals.

Calculation of the curvilinear integral over the coordinates of the functions assigned to the curves , be reduced to the calculation of the first integral in the form

how crooked given parametrically
showing the cob of the curve , a
- Її end point.

Calculation of the surface integral as a function
, marked on the double-sided surface , add up to the calculation of the underestimated integral, for example, mind

,

like the surface , assigned to equals
, uniquely projected onto the plane
to the region
. Here - cut between a single normal vector to the surface and all
:

.

Consumed by the minds of the head side of the surface is determined by the choice of the sign of the formula (2.3).

Appointment 2.1. vector field
is called the vector function of the point
at once from the area її assigned:

vector field
characterized by a scalar value - divergence:

Appointment 2.2. flow vector field
through the surface is called the surface integral:

,

de - a single normal vector to the selected side of the surface , a
- scalar doboot vector_v і .

Appointment 2.3. circulation vector field

on closed curve called the curvilinear integral

,

de
.

Ostrogradsky-Gauss formula set a link between the vector field flow through a closed surface and the divergence of the field:

de - Above, surrounded by a closed loop , a - Single normal vector to the surface. Directly normal, there may be benefits from a direct bypass of the contour .

butt 2.1. Calculate the surface integral

,

de - Zovnishnya part of the cone
(
), which is seen by the plane
(Figure 2.1).

Solution. on top uniquely designed in the area
flats
, and the integral is calculated after formula (2.2).

Single normal vector to surface we know from formula (2.3): . Here, for the normal, the plus sign is selected, the shards are cut mid-air that normal - stupid i, otzhe, may be negative. Vrakhovuyuchi sho , on the surface acceptable

Region
є kolo
. Therefore, in the remaining integral we pass to polar coordinates, for which
,
:

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

Solution. For formula (2.4) we take

The rotor of the vector field is known from formula (2.5)

butt 2.3. Know the value of the vector field
through part of the area :
, roztashovanu at the first oktantі
).

	Solution. The force of the formula (2.6) . We represent a part of the area : , raztashovanu at the first octant. The alignment of the given area at the windbreaks may look (Figure 2.3). The normal vector to the plane can coordinate: , single normal vector
	. . , , stars , otzhe,

de
- area projection on the
(Figure 2.4).

Example 2.4. Calculate the flux of a vector field through a closed surface , settled by a flat
that part of the cone
(
) (Fig. 2.2).

Solution. Accelerated by the Ostrogradsky-Gauss formula (2.8)

.

We know the divergence of the vector field formula (2.4):

de
- obsyag cone, yakim conducted іtegruvannya. Speed ​​up with the home formula for calculating the volume of the cone
(- radius of the base of the cone, - Yogo height). For our mind, we can take
. Residual

.

butt 2.5. Calculate the circulation of the vector field
along the contour , covered with peratin on top
і
(
). Verify the result using the Stokes formula.

Solution. Peretina zaznachenih surface є colo
,
(Figure 2.1). Directly bypassing the vibrate, sound those schob surrounded by it, the area was left with evil. Let's write down the parametric alignment of the contour :

stars

moreover, the parameter change in before
. Behind the formula (2.7) from the equations (2.1) and (2.10) we take

.

Let us now state the Stokes formula (2.9). Yak surface , stretched on the contour , you can take a part of the area
. Directly normal
to tsієї surfy zgodzhuєtsya z direct circuit bypass . The rotor of the th vector field of calculations in application 2.2:
. So shukana circulation

de
- area of ​​the region
.
- near the radius
, stars