Ders çok değişkenli fonksiyonlardaki ikili dizinin birincisidir. Burada çok değişkenli fonksiyonlardaki temel türev ve entegral kavramlarını geliştirmek ve bu konulardaki problemleri çözmekteki temel yöntemleri sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Genel Konular ve Düzlemdeki Vektörler Bölüm 2: Uzayda Vektörler, Doğrular ve Düzlemler; Vektör Fonksiyonları Bölüm 3: Düzlem Eğrilerinden Hatırlatmalar ve Uzay Eğrileri, İki Değişkenli ve İkinci Derece Fonksiyonlar ve Karşıt Gelen Yüzeyler Bölüm 4: Özel Yapıdaki İki Değişkenli Olarak Karmaşık Fonksiyonlar, İki Değişkenli Fonksiyonlarda Kısmi Türev ve İki Katlı Entegralin Temel Tanımları; Limit Kavramının Gerekliliği ve Anlatımı Bölüm 5: Türev Hesaplama Yöntemleri Bölüm 6: Türev Uygulamaları Bölüm 7: İki Katlı Entegraller ve Uygulamaları ----------- The course is the first of the sequence of calculus of multivariable functions. It develops the fundamental concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Chapters Chapters 1: General Topics and Vectors in the Plane Chapters 2: Vectors in Space, Lines and Planes; Vector Functions Chapters 3: Reminders of Plane Curves and Space Curves, Quadratic Functions and Variables, Surfaces Chapters 4: Special Two Variables Complex Functions, the Basic Definition of Partial Derivatives and Two Storey Integrals in Two Unknown Functions ; Necessity and Details of Limits Chapters 5: Methods of Derivative Calculations Chapters 6: Application of Derivatives Chapters 7: Two Storey Integrals and Applications ----------- Kaynak: Attila Aşkar, “Çok değişkenli fonksiyonlarda türev ve entegral”. Bu kitap dört ciltlik dizinin ikinci cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 3: “Doğrusal cebir” ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Calculus of Multivariable Functions, Volume 2 of the set of Vol1: Calculus of Single Variable Functions, Volume 3: Linear Algebra and Volume 4: Differential Equations. All available online starting on January 6, 2014
Dairesel Koordinatlarda İki Katlı Entegral ve Çözümlü Örnekler
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En provenance du cours de Koç University
Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts
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À partir de la leçon
İki Katlı Entegraller ve Uygulamaları
İki katlı entegrallerde hesaplama örnekleri. Kartezyen ve dairesel koordinatlarda hesaplamalar, uygulamalardan örnekler.
Rencontrer les enseignants
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
In the previous session Cartesian coordinates rather
comprehensive of two layers integral We have seen how to do.
There were some interesting details.
We have seen that the order is important.
If done properly accounts in which you do the same integral
With the appropriate choice has consequences, but more we can conclude from the short.
I have seen them.
Several of the variable edge, variable
how to handle the limited areas We learned.
Now here from the Cartesian coordinates after
double decker circular coordinates integration will do.
Why is this important?
Many can be found that a long list.
But most of the following two points.
Once in nature and technology appropriate self-
emerges as a circular geometries.
This is important because we see these kinds of issues need.
For example, the beginning of civilization wheels We think that it's a circular shape.
Even if the circular itself is a good
according to the model building approach structure.
In his circular coordinates We need to learn to work.
Please note that the circle of civilization, the wheel with the discovery
There are many who have started building.
In Cartesian coordinates x delta delta than y
with an infinitesimal area element comprising We would begin.
Now let us remember how we were doing.
We choose an x, there is a vertical line We're taking.
Here again as an increasing delta x We're taking a vertical line.
Similarly, we choose y.
A second line from the y plus delta y We're taking.
The intersection of these four lines us here you can see
infinitely small area of the rectangle as giving.
The same thing in circular coordinates 're doing.
We choose r is a constant, how the where x represents a constant chose.
R is fixed, but a circle, radius r the circle.
We are expanding it a bit, how x is a We have increased the amount of r plus delta
from r, and that of a circle, the second circle We gerçiriy.
This of x from x and x plus delta lines are equivalent.
Similarly, we look at the second coordinate.
Our second coordinate theta.
Because any point in space at the center distance,
the distance r and the angle from the x axis are determined by theta.
We've seen this before.
Therefore, we are having a line of theta.
Theta is equal to the mean constant.
From this center by angle theta 're right.
You are expanding the delta to theta.
Therefore, the theta plus delta theta We draw.
There are four lines as you see here.
As in the same Cartesian coordinate here we get a rectangle.
But this is the curved sides of the rectangle.
But that does not matter much.
The scanned area of the delta is important finding.
Find infinitely small space.
It can be calculated in two ways.
One, working with infinitesimal method
For this rectangle, line This rectangle
The difference between this edge with the edge of the second is of the order.
Burr is the length times the delta t.
This place is r times the r plus delta delta is theta.
Thus where r is the edge delta where r delta theta theta should get it.
If we multiply the length of it in the other edge is the delta because it is a circle in
In the circle r r plus the delta, as you can see
the infinitely small area directly we find.
Here is equal to the symbolic value, we obtain d theta.
Deltas turning to d.
If you are not happy with it, as well as a educational
There side, precisely this area Let's calculate.
This way the absolute calculation area As these are
r and r plus the delta between the circle Let the remaining space.
We pi times the radius squared area We know that.
So it remains in the outermost circle Once the area is pi r squared plus delta.
This time the inner circle is pi times the square.
As you can see by these two, it emits a r opened the next frame, the frame is
Once the delta is going to be run twice The term here.
And the delta will be the square of r.
This was the area of the entire ring.
We are the delta area of the angle in radians We want to find.
Its a very simple thing to do.
This area of the ring space the two pie If we divide the area AndAki'll find a raft.
This delta t and delta theta with
delta theta angles opposite impinged We found the area.
Here are two pie take this ring area I divide
where p is going to like what you see.
Here comes the bottom two.
And I'll stand tetayl delta.
When we organize these terms, how We organize,
If we take out the edit delta r r
next to it is a plus an infinitely small There are a range of terms.
It symbolically to the infinitesimal One glance is that the delta near the negligence
will be back for the fall and the same short way, these results are obtained.
I see a tutorial on this side because upper
order that an infinitesimal We see that next fall.
This will always be the case.
Beeline for it when he calculated that we automatically As lowers itself would have.
With the difference in length because it broadcasts
the outer circumferential length of the spring is the difference between.
As a trapezoid if you think that the outermost plus the inner edge
he divides the two sides of the same thing if you receive is divided by two to one you would find the difference.
The result means that in terms of infinitesimal The sum of the area of the delta.
Delta integral to the sum of is coming.
This is where, dr, d of theta coming to integration.
To a two-storey integral.
There are two variables r and theta.
Because I find the location of a point in the plane r the distance
the x-axis by the distance r to the center the angle made by the two variables.
Here again, any point locate
Two variables are needed for the x and y.
Select a sample immediately after it do it.
The area of a circle with a radius Let's find.
Of course we know it since middle school.
We can do this in two ways.
Also in a circular coordinates in Cartesian coordinates.
See the outer circular coordinates ring, R is A great circles.
Inside this small area of Delta we see.
When we do this account r Reset boundaries between
is changed, reset the two theta pi We see that changed between.
So while it is integral, is,
is obtained by typing the limit theta We are.
But that constant values for these limits
and integration consists of a single function, r.
Write once he r r r times of dr'y As an integral, one-ply
integral of the theta d theta a hit In the second integral we find.
See here r squared divided by two turns.
The upper limit is happening when you put a squared divided by two.
This is also the theta theta integral is happening.
The upper limit of two pins, the lower limit is zero.
Here comes two pi.
Simplify the work of each other Twos again from middle school
obtain the formula pi squared, since we know We are.
As you can see supremely simple I have found accounts.
Because the circle, the circle the most appropriate for
natural coordinates circular coordinates Because.
In the same account in Cartesian coordinates I give homework to do.
As you can see here a great deal more accounts would be long.
This account is very simple geometry for can
but in any case relatively long would calculate.
This is because the linear coordinates, Cartesian
coordinates correspond to a circle coordinates are not.
Here why we thought it was easy,
A very simple definition of a time limit decreased.
R is A.
It's that simple.
But here in Cartesian coordinates limit is not so simple.
Moreover, the double-decker integral variables separated
integral to the product of two one-storey has become.
We were a little more of our business
ilerletsek and more circles, circle we find in trying moments.
Moment was very simple.
Wherein the domain, the infinitesimal area, We were looking to their distance from the axis.
From the x axis of these infinitesimal area from y.
If you get a point in the center of it distance y.
Ha this is infinitely small for them You bought the corner, huh
've received somewhere in the middle, huh edge 've received, all for going to the limit
the infinitely small area to zero, a point will be in the range
Let me HERE, HERE, let me There's no need to think.
y at once.
But we know in Cartesian coordinates a point
x-coordinate is r times the cosine of theta.
We can write it brought.
Similarly, y is the r times the sine of theta.
When we place them see y r instead of sine
There is also another one in theta, hence r squared sine of theta is happening.
This integration comes to the account.
When received by the first moment of the y-axis See there again
but seemingly foreboding clash x length logic is right there.
So in my x's, y's mx.
It previously also in Cartesian coordinates 've seen.
We are writing instead of x r cosine theta.
There is also another one in the that are Square is doing.
is integral with two floors as d theta we find.
From these centers of gravity We learned that.
As for the second moment still have the same too.
x and y are expressed by the same transformation.
According to the second moment of the x-axis y
Once an area of the square is an integral knew.
y r squared sine-squared theta will also square.
There is also another one in, r is d theta.
As you can see, this expression occurs.
similar iyy the same thing.
That is to memorize something new in mind, do not need to keep.
Ew, that the square of the distance from the y-axis contains.
You're writing in Cartesian coordinates.
Bring this circular coordinates expressed in We are.
Ixy'y also told, in this way A measure of deviation from symmetry.
Would be important, for example, on a triangle We have seen,
IXY and sometimes not, sometimes.
In which x and y are also IXY
ie the integral of the product on the field learn from them
Even for those who want all of you with them greeted
that is an integral number of them Examples for the calculation creates.
Our ability to calculate this integral In order to improve.
But it is also very useful information as well as We also broaches obtained.
As you can see here sine cosine theta theta is coming.
Yet here is because there is going cubes, a
There is also a drink here than there are r cube is going on.
In the center of the field is zero, this infinitesimal the square of the distance.
X squared plus y squared he also referred to as We were not.
And x squared plus y squared is the square.
Thus, here comes an r squared.
Coming from a run in, as you can see is made extremely simple terms.
Now let's implement it.
We have the circle's area accounts but behold a Let's see once again rapidly.
And it is in the area moments, dr d theta is this
as the two floors of the integral We need to marshaling.
It does not matter what order you get here because there are limits on the constant.
There are also limits on the hard theta.
These functions that can be separated theta by observing their
We can calculate the integral in, r We can calculate the integral in itself.
Here two pi, so a square divided into two As we know, we find the formula.
The first moments when we account again
From y in the same way a sine theta sounded.
Here the integral of x sine of theta
meant cosine theta, cosine theta integral.
Here, too, the variables can be divided into integration.
But now we see that the sine of theta two pie from scratch, so the integral is zero
because for a period of sine theta
If you receive negative areas plus areas will take.
If you want to analytical cosine theta happens.
A cosine theta values in two pie.
A zero value in the, apart Because you can remove is zero.
Here, too, the cosine of theta integral If we take the sine of theta happens.
Sine of theta zero in two pie.
Zero value at zero again.
Zero minus zero is zero.
Therefore, centers of gravity zero coordinates given by
and we also know it already, a circle The center of gravity is the geometric center.
Therefore, it has provided our account We're going.
Have confirmed that we are old words.
If we look at the second moment again ixx't You do not have to remember anything.
Mean distance from the x axis second moment of x away from the axis.
We measure the distance from the x-axis and y.
Here is a sine-squared square next year, is the integral sine-squared means.
Here is the next square, in a run in the There is, for that cube.
This two-fold integral with large open we can do.
Minus one-half times the cosine two theta.
Two theta cosine decreases, here of a p income.
This pin is the integral of the cube is also a four divided The four data.
As you ponder that the moment of inertia we find.
y-axis by the same moment of inertia
things here because the integral of the square cosine is there.
The integral of the square of the cosine of the sinus is identical to that of the square.
From there, a pi comes again, there is also a four income divided by four.
As you can see here IXY sine theta There are times the cosine of theta.
Because it gives zero integral sinus makes two theta.
Sine the integral of two-theta plus
areas will take the field minus zero happens.
We have learned this already happening, He remembered the
we are, though IXY symmetry deviation is different from zero.
Which is completely symmetric in the circle, x and
According to the y axis, it will be calculated as zero are natural.
We're looking at is zero.
I will remember at zero sine-squared plus the sum of the square of the cosine of the future.
Ixx and AOP.
And as you can see here there.
So is there theta integral.
Here come two pi.
from the integral of the cube is a four divided by four future.
But with two pins in two of the four divided by four two
sadeleşerek get pi divided by two to four It is.
In Cartesian coordinates to make them even advice
I'm not fantastically complicated because will take you to the account.
Not be done, but it's simple, clean
quickly achieve the results not possible.
Your homework in a semicircle field and
I suggest finding the torques.
Of course, we know that half of the field.
Here to guide you as the full apartment zero theta limits
While wandering the two pie in a half circle just come up to scratch pie.
Therefore, all those integral reset of p we
When mx, ie, x-axis
According zero moment will come.
Because when we receive full circle following and
the upper half circles will balance each other and has zero interest.
This will go to zero.
But that the left and right semicircle
to balance each other as parts of my will be zero.
From here on the x axis already coordinate of the center of gravity
We find zero because the area my'y will divide.
But this half-circle in the y direction
will be zero because the center of gravity There is nothing at the bottom.
And it is a four-thirds pi account will exit.
You get it by taking away space divided into I would anyway.
This is roughly half of radius slightly is becoming less.
IXX, Ew, wherein the values IXY is given.
When you do these accounts of right and wrong Do
're doing, to be able to direct you to a simple calculation.
A second example of the space curve finding.
Bell curve, also encountered in probability the Gaussian function.
Minus one-half of that override the minus the
areas that may be something a little more A practical, a more appropriate structure.
At this function, probability calculations and heat equation, heat conduction
frequently in the solution of equation is a function encountered.
Now we want to find the area.
Although this integral to the minus x we can do but to the minus
coordinates of x squared term, since it is It is not possible to do with the transformation.
But it is integrally closed integral integration can be done by turning on two floors.
For this, such a simple idea 're doing.
to square minus one-half years of dy'n the hesaplasak integral y
At the end of integration will finally fade away for
calculated on the same integral with x It is the same value.
This is the II.
In the first, as I have defined.
Therefore, if we multiply them, see There are x squared and y squared terms.
Where the integration, where the integration.
I will give this square.
Now we will gather these two integral.
Let's get together.
Of course a negative exponential collected and divided two, to x squared plus y squared.
Now we see that x squared plus y squared r The frame data.
Here is the area dx dy, it also gives.
We arrived in circular coordinates r, dr, d theta.
We know it.
So here it is I'm putting off the range tetayl are seen as two different terms.
Now this again with separable variables kind of integration.
Because here only the functions r is there.
Has only one theta functions.
Now, what happened, we all in the xy plane from minus infinity to x on the plane
plus infinity, minus infinity plus y I have found taking forever.
To do it in circular coordinates path such
You can get an apartment from scratch, as r goes.
If you also take an infinite plane r all We have been close.
However, due to the hippies from scratch by theta from zero to infinity as r
so, of course, is the meaning of minus infinity No, negative values.
it has a length.
Length is the surplus value.
Already from the center to the edge of the circle one.
We also forever a circle of radius We are going to get from zero to infinity.
Integral supremely easy.
The first integral we find two p right away.
The second integral is also a little thinking we need to do.
An integral single storey landed.
But this time, the following is our business to facilitate.
As we do, because a change of variables, minus one-half
r squared to say u, d refer to the two lead here this fall
is simplified by two negative r would still is.
When we took it replaces See above where e occurs.
the other is made from dri is a minus with a difference.
The cons so.
Now u r is zero is zero.
But when u is negative infinity is infinity happens.
They also borders also are organizing.
Now this can be an integral integral has become.
E u e to the integral over u.
U equals minus infinity to zero boundary We're taking.
Now we see that e to the minus infinity minus infinity is zero, falls.
If zero, equals zero if u is a happens.
So we're finding zero minus one.
Minus one involved here.
But at the beginning there is a minus.
Thus, we find the two pi.
Already have an added value still logically get an idea as superficial
plus valuable because it can be anywhere This means that the sum of a function.
Or it means that the area below.
It is clear that the pros go.
So I squared equal to two pita our Our integral is equal to the square root of it.
This single-storey otherwise by integrals I can not find.
It genellenerek slightly different places
You may have seen it, here one sigma are placed.
Put a sub in here one sigma sigma comes with coordinate transformation.
This gives us an integral again.
This indicates a probability distribution so in many applications.
Or heat dissipation issues such as a plane from a point
ısıtsa the bell curve that after a while you takes shape.
These are not our business, but a lot of them to efficient
to tell you that there are functions I give this example.
Already now it has become fashionable to content I explain math.
Of course, this time is very exaggerated mathematics öğrenmeyip
physics, statistics, doing the economy is doing deduct state.
Him from falling, but their relationships There is also the psychological value of knowing.
You can see your other courses possible relationship issues immediately.
The function of an upper limit of the variable By this variable, so
integration variable x base a saying We can obtain new functions.
Because here in the integration of x base will go
Place of x will be a function of x.
It us up to now we have not encountered of exponential functions
The algebra in a more upper-class exponential functions
as a function of the family the opportunity to know giving.
It is frequently encountered, however, said such as the
probability calculations and heat transfer in problems encountered in kind.
We then see a new 'll pass.
With the circular coordinates at a start.
In the second part of this course is more comprehensive As we'll see.
Here are a few very basic do we deal with.
It's a little more deeply, and some more the acquisition of skills required field.
But here I've figured it out for two reasons, albeit briefly
circular coordinates without once I'm bitirmeyel this course.
You'd make a unity in itself.
Some of the second portion to discontinue you may think.
They would have seen at least the.
But also encouraged those who want to continue to
it is a tip of the iceberg I want to finish by saying.
Because more circular coordinates There are many important issues to be seen.
After that, we will write a new chapter.
This again is very comprehensive, is a matter of deep and supremely well is an important issue.
Because of the nature of this general equation are obtained by application of the gradient.
We'll make an entry.
For the same reasons that the course of the gradient To finish applications without
For an idea because it's so simple too will be met with a field rich.
But on the other hand, this rich area where If you see the second part
would be an incentive to continue I think.
For him it albeit briefly, this rich We will recognize the subject.
For now, I'm finishing this course here.
The end of this semester, this course Or
group, have come to the last part of this volume We're going.
This gradient and their applications
nature, seeing how he got to the equations you'll end up.
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