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
Düzlemde Kartezyen Koordinatlarla İki Katlı Entegraller
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Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts
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İki Katlı Entegraller ve Uygulamaları
İki katlı entegrallerde hesaplama örnekleri. Kartezyen ve dairesel koordinatlarda hesaplamalar, uygulamalardan örnekler.
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
In this order in the previous sections We watched.
After a general introduction in the fifth chapter multivariate
and the definition of partial derivatives of functions In the same section again multivariate
integration of functions in multi-storey We have seen definitions.
One section in chapter five short applications are also made.
Six of the seven and eighth chapters of this what we have learned
we make the calculations more practical sections.
In the sixth chapter in the calculation of derivatives I have seen methods.
In the seventh part in derivative applications 've seen.
In the eighth chapter in this section
integral calculation methods and We will see applications.
Derivatives were separated in two parts because it
As pretty as a method of calculation issues had.
For this reason, we have divided.
Apps in a separate section, We have compiled.
The purpose of this calculation section our ability to develop the same
In time we learned theories ensure consolidation.
We'll start with a two-storey integral and We will see in a plane before integration.
They are in Cartesian coordinates and will be in circular coordinates.
Watching this part of the course that it is very variable
differentiation and integration of functions of the first We will have completed the skin.
But the issue is not over.
Integration of the main issues we shall see
Following the integration we see here We will see in space, surface integral.
So here now two floors in the plane integrals in space, while a
If you think it's two-sided surface two dimensional body.
He will bring a two-storey integral but
different information to be had in this space need.
In space on three floors in the same way again We will also see integration.
In this second volume of both of these programs What more
to introduce'll see, maybe a little to advertise in
but I'm telling you this for a joke, since each This work does not stop here alone
When you finish the course by a lot of basic information 'll be able.
A robust infrastructure that will be formed.
Then you will bring.
Now I'm starting with something like remembered.
Integration of functions of one variable As the concept was born.
Our f x as y is equal to such a curve get.
Under the curve from A to B until a area are saying.
The way to do this infinitely small accounts study of the most fundamental approaches
one of the finite range of b eternal we divide into small spaces.
That each interval under the curve field with a value of approximately reckon.
When collecting them about the area We find a value.
Watching it when we get the limit all these deltas
go to zero when the derived We call the integral limit.
This we have done.
In bivariate transactions and fully developed concept
but of course that similar equivalent be generalized.
We know that z is equal to three-dimensional space FXY
that is a function of two variables showing surface.
In the plane of the area under a curve Our finding of two variables
this surface in function of the anti- find below the volume.
To do this, our approach is exactly the same.
How infinitesimal interval in b We were split into pieces
where x is of course the year as well is there.
Therefore, an infinite range of x is a b would divide into small pieces.
this time in the year forever c d range we divide into small pieces.
This we have achieved a small rectangle
f x and y, we find from on high.
It would stand with the base area.
On both edges of the base area length delta x
and that the floor area, including the delta y.
Separately with the function where it gets hit as
We find infinitely small volume drawn.
They are also about the time we collect as
The volume under the curved surface we find.
When they go to the limit delta x and delta y of the limit
This time when we went on a two-storey integral're getting.
This is a natural development.
An integral one hand while collecting We are.
This time while collecting two two-storey I need to achieve integration.
At this than you are already an integral icon
I've said before, English or in German
water or initials of the words summa collection within the meaning of
pull the first letter of the words of s elongated shape.
Of course we are collecting year ago
on the x's on
As we do on the first x over the years as we can.
We have seen a fundamental theorem.
Simple logic tells him already.
Want to infinity before these small volumes Let's move in this direction by collecting.
I'm picking up the y direction after sonuçlandıral or
Let's go in reverse order that the same result is open.
So the order of the terms in a collection changing're collecting.
The calculation of the basic approach while
functions of several variables in the basic
that is, a variable approach to fixed
We keep changing the other two variable.
Forty-nine to fifty if our variable We keep constant temporarily, of course.
Hence one would change we know the problem
functions of one variable structure We bring temporary.
Partial derivatives were doing the same process again We're doing the same process in the integral.
Now here's a simple example to start I want.
Of course, more complex calculations methods, but we'll see
Once you understand the main idea at all close down.
a'yl of x b y c and d of the is
x y plane upon which time course There built surface.
We can do the integral over the prior year As this one strip
that means taking on the integral We choose a fitting where xi temporarily.
After receiving this integral, after counting After this
second time on the remaining variables We calculate the integral.
It also could make in the other.
Equal to the simple logic of the two theorems would say.
Our region is always a lonely
Simple as defined by a rectangle may not be.
For example, with constant values, where x identifying
y direction, although the curvilinear boundaries present.
One year ago on the appropriate approach when is to calculate the integral.
X on it because we would calculate We will see in more detail.
It's hard to see here on the surface curve coming
but when it came to the central region of constant comes hard.
When you come here again without a boundary curve We're going hard again.
This happens because a suitable three different region requires.
Similarly, if y hard this time appropriate
x over y is held constant as before because change
regions where a single type of You can define the whole region.
Conversely, if we had done in the other See until the point that broken
y has a fixed value of a variable value From this point, the second fracture fracture
Y is a constant until point to a constant value and the last
Y is a fixed value as the curve would be coming to a limit.
This of course would not be an appropriate way.
Our three different integral hesaplamama should
We were supposed to do three different collection.
Let's say the two curves in some regions between these varieties
whether this way before in a region y sabitleyip
are calculating the integral over x or x y on sabitleyip
integration are calculating the same difficulty or because it will be the same ease.
All in a single site definition We can cover.
Wherein these functions more As I've written.
For example, when we secure ago x We are moving along the vertical means.
G of an x, y is the vertical g for two xA is reached.
After doing this account at this time x, x fixed on
a value from one end to the other end, we're going.
In all these various applications We'll see.
Others may think of this time to get y We will hold x on a fixed integral.
But here is the function y equals h us he quoted.
In contrast to fix these functions ie we need to take.
Where y g a x
Given this inverse function as it x to y be denominated accounts.
X is H, such as that the equivalent of one year function comes to be shown.
In some cases such a circle ellipse
the region, such as a continuous curve, closed curve can be given by.
Here also artificially limit the cut here you have two options.
Before you can fix the smallest x x largest value
artificially around until the value of x an interruption
thus bringing the curve y equals
g g y is one and two the two expressed in parts
y or the equivalent to the When this sabitleyin
time here, bringing an artificial cut x this h a
value until two values h would be developed.
A typical example of an equation for a circle of course if you receive
a circle, a one year against every x no.
When we opened it for her a circle
two of these closed curve representation income counterparts.
Here y is equal to minus the square as if cut
plus y equals the square root of the term is rooted y as much.
In terms of the equivalent account the x-y You can.
Where x is the value minus the square rooted
the second branch plus square rooted values would like to take.
Now let's start with an example.
Let's start with an example from the most simple.
Whether a function is as follows:A feature No this function and a rectangular
Let this integral accounts within the region.
So this is the remaining volume under the surface
formed on top of rectangular volume calculation comes.
x from zero to two years less a merger of such a rectangle.
As I said just before x hard y on hold
We can do the integral or y is held constant x on the integrals can begin.
It is clear that both give the same result.
've Done it before, this kind of simple
For an application, but to remind here I'm going out again.
In this example, the order of integration is not important.
But we both want to do.
The first step is held constant over x integration.
Here are two integral on a field given.
Now it becomes more regular functional We need to bring.
Before we hold constant y on x will be integral.
During this process x temporarily fixed happening.
Thus if we account this integral for now
Do not ever consider the integral of x on.
Four minus x squared is making a hard task.
an integral brings y on y.
There is a square where a year from its integral y cube would be divided by three.
Limits are putting a negative and y is equal to as y is equal to one.
Yet at every step of a single storey integrals We're trying.
Put these values, y will go.
Because we now know the values of y.
When we put this year in 4 minus x squared A minus minus one year, minus
year divided by the cube of a cube in three years a year, minus one minus the cube.
This is that.
Terms of these terms, we edit it The detail is very simple.
Currently, this term only in terms of x obtained We are.
Now that we know the integral single storey.
This integral we make an account of this first fixed term
hard times x in X, because of the constant of integration is who is possessed.
X in the second frame.
the integral of x squared the x-cube divided by three.
But in the beginning there were a few.
X is equal to zero, and this brought the two the values are calculated.
When I made this account, the x is equal to two If you put the future of 44 here.
When you get to the next 16.
When this account is going well below 28 There are three in the denominator.
This is the problem we finish.
However, to satisfy ourselves thoroughly this theorem for a
To see provide at this time in many different ways, let integral.
Here again, the general two-storey integral written.
Now when you type it on a regular basis We take the hard x variable y.
This integral again in the second integral We forget temporarily.
Because a hard year for us now.
Four minus y squared.
This is a constant of the integration thereof of X, who is possessed of.
There is a square integral of x squared minus x the x cube divided by three.
Account the values at zero and two We are.
Below is zero at ease a little brings.
Dropping all terms, of course.
When we edit it minus 16 divided by three two years, staying square.
Here is seen the details.
Now we know this gene is a single-storey has become integral.
When we calculate this integral divided by 16 had three.
This will give the integral y, to y'yl who is possessed of the square minus y
They are integral y cube divided by three minus one and in
When accounts for 16 minus two 14.
One of them the same value minus one Because it was still going 28 divided by three.
Before we do this in many different ways integrals give the same result.
Therefore, we have provided this theorem.
And again we did in the fifth chapter
An example of this is actually to remind 've done.
I'm doing here.
I also separated variables we can do.
I'll leave it as homework.
If x and y are rectangular because of constants and functions of
wherein x and y can be withdrawn from the genus you can not separate all integral.
But only one can separate the integral.
In this way, the same result if we again We'll find.
Also to make it to you as homework I'm leaving.
Now a very rapid and simple integration We have reviewed the issues.
A method for making integral various that region.
Now we take a break.
When you get a thought.
They get the practice time.
Then more seriously now more complicated but it is not difficult to regions
easily be realized once the system We will give examples that can be done.
Goodbye.
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