Ders çok değişkenli fonksiyonlardaki iki derslik dizinin ikincisidir. Birinci ders türev ve entegral kavramlarını geliştirmekte ve bu konulardaki problemleri temel çözme yöntemlerini sunmaktadır. Bu ders, birinci derste geliştirilen temeller üzerine daha ileri konuları işlemekte ve daha kapsamlı uygulamalar ve çözümlü örnekler sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Multivar 1'in Özeti, Dairesel Koordinatlarda Entegraller Bölüm 2: Türev Uygulamalarından Seçme Konular Bölüm 3: Çok Değişkenle Zincirleme Türev ve Jakobiyan Bölüm 4: Uzayda Yüzey ve Hacım Entegralleri Bölüm 5: Düzlemde Akı Entegralleri Bölüm 6: Düzlemde Green, Uzayda Stokes ve Green-Gauss Teoremleri Bölüm 7: Stokes ve Green-Gauss Teoremleri ve Doğanın Korunum Yasaları ----------- The course is the second of the two course sequence of calculus of multivariable functions. The first course develops the concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. This course builds on the foundations of the first course and introduces more advanced topics along with more advanced applications and solved problems. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Bölümler Bölüm 1: Summary of Multivar I, Integral in Circular Coordinates Bölüm 2: Topics of Derivative Applications Bölüm 3: Chain Derivatives with Multi Variables and Jacobian Bölüm 4: Surface and Volume Integrals in Space Bölüm 5: Flux Integrals in the Plane Bölüm 6: Green in Plane, Stokes in Space and Green-Gauss Theorems Bölüm 7: Stokes and Green-Gauss Theorem and Nature Conservation Laws ----------- 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 Çözümlü Problemler
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En provenance du cours de Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
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Koç University
9 notes
À partir de la leçon
Multivar 1'in Özeti, Dairesel Koordinatlarda Entegraller
İki değişkenli fonksiyonlardan hatırlatmalar: ikinci derece fonksiyonlar, kısmi türev ve iki katlı entegrallerdeki temel tanımlar ve geometrideki anlamları; iki değişkenli fonksiyonlarda türev ve entegrallerdeki temel hesaplama yöntemleri, iki katlı entegral hesaplamasında sıranın öneminin örneklerle hatırlatılması; teğet düzlem ve diferansiyel; tam türev ve zincirleme türev. Yöne göre türev. Gradyan, Bu sonuçların üç ve “n” değişkenli fonksiyonlara genellenmesi.
Rencontrer les enseignants
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi there.
Our previous problems that circular
coordinates here circular coordinates.
However, a significant difference.
Before this we solve the two problems r is the line of constant values??,
theta values ??of the boundaries were hard.
Theta is between two pins at reset In a first embodiment of resetting,
In the second example also reset was among endless.
Now we will see here in the problem limit will be fixed at all.
Examples will make significant examples.
One of two types of maintenance skills I hope that you will win.
To internalize the circular coordinates,
In these coordinates with infinitesimal To write the integral thinking.
Our first problem radius a and center at x is equal to
and y is equal to zero, the We want to find the area of ??the circle.
Of course we r and theta is There are two options for.
Before we can take the integral over r.
Before we can take the integral over theta.
Now we will see that both What is important to formulate them.
The result of these problems, of course, obvious.
Where is the area bounded by a circle Whatever is pi times the radius squared.
Therefore the answer already We know that pi squared.
But it is important that we know the result here In the example in b be able to write properly.
Do not know the result, then a little
A sample of air is going to stay I chose this particular way.
Now before this circle E,
circular coordinates Let's terms of software.
In Cartesian coordinates this circle equation x minus
because the square of a year, the center plus
because the central square in terms of y y y is equal to zero is equal to a square.
When you open it, see from the first term future of x squared.
y frame already exists.
But the first term of the square a future minus two times x.
BI will also squared.
But it has left the bi squared.
There is a square on the right b.
Therefore, they mutually and simplify the right side is zero.
Now this equation, in Cartesian coordinates this
Let's write the equation in circular coordinates.
We know that x squared plus y squared is r squared.
Minus two are writing him.
x If we saw earlier, wherein an angle of the point k
When the theta x component, x-coordinate is r times cosine theta.
We are writing the cosine of theta is once.
As you can see here, there is an extra b.
r is zero sensible solution We can not reduce it to that.
Where r remains.
A glance at the right side of the two times the cosine of theta be.
Because it is here is more simple.
Now before holding constant theta Let the integral is over.
d r d r d theta know that.
So if we do the integral on first run What will be the limits of r here?
As you can see is a theta constant We are moving on a straight thick.
r is zero, this Up to the border.
This border is well worth EUR equals two a cosine of theta both a cosine of theta.
We are also writing.
So it became clear that the limits of integration.
When it comes to limits on theta As we can draw this circle.
Here begins as follows: If you go negative pi divided by two.
Here increasingly smaller angle, or else is smaller in absolute value.
Theta is equal to zero is coming.
Then when we move from here theta
When we came here is growing is growing pi divided by two is going well.
So the boundaries of theta constant values ??but
r from the border of one variable bi.
This integral is easy.
Because here as a constant theta We think operationally.
r r squared integral divided by two.
We will put in place When a square is going on four.
But there are two r squared divided by two from the bi hence the two squared in the denominator is coming.
Bi square cosine theta.
Of course, because it is a constant cosine square outside inside.
General standards of doing this integral way to use the double angle formula.
Cosine squared theta minus sine squared theta We know that the cosine theta equals two.
A minus sine cosine square frame we make the square cosine of theta
divided by the cosine theta plus two We both know that.
Here too there are two bi.
So the dilemma is gone.
Had only a square outside.
Limits are the same.
Instead of a square cosine theta cosine theta plus two arrived.
Integration of two cosine theta zero.
Because of this integration in the region we make two sine theta.
Two sheep instead of theta pi divided by the sine of pi.
E sinus p is zero.
If you put the lower limit sinus minus pi, it is zero.
Another optically,
two more practical geometric optically cosine theta range, this field is zero.
Because valuable space plus minus valuable takes you back to the fields and zero remains.
So the cosine of theta integral falls.
Only one here is the integral theta.
He also divided the two above minus minus pi pi
therefore two divided by two times pi divided by two is happening.
Pi is happening.
Here there was a square means that pi squared.
This result is already known.
But here, I think both Thinking in terms of both
Ability to write the account limits I've got an important win.
Repeat it here.
t we fix the case for b're going on the right.
wherein when the value of r is zero it comes to the value on the curve.
Variable values ??on the curve.
Where we found different.
Depending on a cosine theta theta two.
You learned to write.
The second integral is a way before we so we got out of theta constants.
The second way bi r held constant theta may be integral to take on.
It appears this second way.
We repeat it here.
r if r is kept constant is part of a fixed circle.
iii eÅiit this is a.
Wherein its boundaries are both a cosine theta
As the inverse function If we calculate a r divided by two.
is divided into two:a.
Cosine theta stayed here.
The inverse cosine theta If we take the function that
a cosine cosine minus one comes here.
That is a cosine function of income.
is divided into two:a.
So we have found in terms of theta r.
Absolute terms of theta values ??are value is the same as the above and below.
But bi income minus theta denominated minus the angle theta here because,
wherein the angle theta.
So that limits in terms of theta
large negative theta functions can be run from the pros theta functions means is large.
Our values ??on this circle integral over theta.
See also integral over r where r is equal starting from scratch.
We draw this circle we draw We are at the end we draw drawing
is equal to two a's coming.
This means that on the r There are also constant.
Now as you can see here, bi We did a complete different situation than before.
We got the first integral on theta.
Then, on the run.
Here's one other tetain We find that it would be theta.
When the negative theta values ??here
This merger shown tert including a cosine function
a minus cosine function function is a variable divided by two.
Above this there has below.
Remove from above so insulted that you b for two-factor income.
You can see the rest of the function only as a function of r.
Now this is very often an integral encountered examples, but these
If we're going to find pi squared integral it is also obvious to us because theorem
in which he says that the integral same day results you get.
But in practice we see Some integrally more
can be calculated easily than some have to be calculated more difficult here
I show here how to do it not an integral impossible until a
This can be calculated by partial integration or take a look at the integral tables.
But in any case this integral is of less conventional type.
The practical purpose of this example as was to emphasize the importance was shot.
Results will not change, but we see that
As in Cartesian coordinates which as also in circular coordinates
Can we get are connected in integral is easy or difficult.
I'm giving you an assignment that
the steps in the previous problems Watching something that can be done.
Here's a semi-circle moments we want.
Particular area of ??the semicircle, Half of pi squared.
I want you to find this moment.
Now to find those moments, one See the difference from the previous example
In the previous example, this angle is negative pi divided Starting from here, two such
pi divided into two parts was going on but in a semicircle pi divided into two expenses starting from scratch.
Therefore, this pi divided Two upper limit remains constant
According to this example, but the bottom limit starts at zero.
Follow the same path of this integral Waiting for your calculations.
Now a second example, see the flag of our country, in
It looks like the moon, but not the same size the inner circle is a circle in an external.
Account to be easily nested In the center of the circle of radius a,
coordinates of the center of the outer circle set in the center of radius of the two.
Now here again Before the fix the theta
We can do the integral on or before r r on theta constants we can.
Refer to the borders before writing Let us try to grasp this logic.
We are between these two circles We want to find the area.
Theta is equal to the constant call We find such a right.
That's right on the bottom Our borders have shifted from the center
will be in the circle of his equation r is both a cosine theta
We found that it will go here until two AA.
When we came here we fix r are We say that such a thing going on.
Here it will expand the r
here that will initially be run at zero then there will be a point increasingly
with growing radius than the the outer radius will come up,
will go to zero, but will say that the two of
they r in terms of the theta function will be the opposite.
The function r is equal to the cosine was Tetala
Assume a wherein R a will be divided into two cosine theta,
the cosine of theta is the negative force,
the opposite of that shown with a minus sign it will be arkkosinüs.
So our problem arkkosinüs will need to function.
Now let's go further this is the first integral on the first scheme
t means that we're doing is secure a cosine of theta, where two outermost
will go up to a value of two, two theta from minus pi divided
here refer to two pi divided come up, but we did it
see here when the circle we're not half the catheter, it also
We need to put an extra integrally As there is a difference here because of theta
When you first see the ring constants does not touch the water's edge just touching.
Therefore, in this second integral theta pi
three divided by two pi divided two will go up.
You're writing the theta pi divided by two from innermost circle is pi divided by three to two, but
r is to touch on the up to a limit of two in going from scratch.
We calculate them See here for a time
Similar to the previous example r squared divided by two, to be released
There dilemma here is always divided by the square
come here two to four frames, come here with four square,
A minus cosine theta future minus cosine squared to give the sine of theta
us and the integral steps In this way we can find out by watching.
Just because we find it to provide without any account we could find.
The area bounded by the outer circle radius
two to four times that of the p means a squared four pi squared,
minus the inner area innermost area of the pin square, that means three pi
and from this integral a squared involved We find again when you have made three pi squared.
We look to the other integral wherein r is the time
See a cosine of theta two two Divided in two to divide a run going.
Here came the cosine theta your ribs Take the inverse function of the cosine of theta
ie the inverse function of theta arkkosinüs function will give theta.
This limit is the equation In terms of theta
arkkosinüs a Tetala function is the outermost
equals two, because it fixed a a full circle here.
If we write this integral the following minus the integral limits
arkkosinüs of plus arkkosinüs going to see here theta
theta where theta plus minus the same There will always be a point symmetric.
It will do the integral over theta it is easy to calculate the integral
're getting one within the limits of the integral There are variables here say that
The above values ??below minus minus arkkosinüs is going the same value twice.
Where r is the integral time easy to calculate, but more
less complicated encountered something integral tables
also be viewed on the previous page As shown this variable
conversion and partial integration If done, this can be calculated.
But again, this example shows it does not fit some sort
and generally circular r coordinates ago
take on a more integral easy integration brings.
This is an example of a spiral, a spiral of drop
a spiral in the plane of the There it is in the DNA helix in space
Or, as in the protein as a Or minaret of the stairs of a house
round like stairs stairs As we saw in the first portion of it that's different.
It opens in the plane that theta r is zero is equal to zero,
theta equals pi divided by two to Here comes the theta equals pi
When this value becomes twice.
When it comes to value, where three times happens, when we came here two
We arrived pie for pi divided by two four times the value of such a thing happening.
See if it's a bird a bent neck
a stylized bird such as the picture of a dove.
Now this is how we will write the boundaries again we have two possibilities.
We keep in constant theta theta ago Hold scratch where r
comes to border variable values We replace a theta theta then,
theta two pie from scratch goes.
When it is summer is d theta Let's statement infinitesimal area.
Limits to scratch a theta going up from scratch.
going to a theta.
Tete two pie from scratch.
Yet it borders One variable.
Gene r squared divided by two turns, is a square frame comes from.
There are two.
Here a square and t, theta is going frames.
Theta integral of the square theta divided by three to give the cube.
From two to zero when calculating this pie involved in the value we found here.
We do the same integral over theta ago.
See if we then take on theta theta is any value we choose,
We start from any border.
Wherein the value of theta We're coming up to this value.
Here we keep r constant.
His first integral on the theta.
From this function we,
r is a function of theta, theta is We find very easily, has a slash.
When this is used as a divided
See wherein
is first divided by a border.
We start from there, is divided into two pie are coming up.
We're coming up to two pie.
This integral is very easy This is one of the integral.
Once d theta.
Theta comes from there.
S of the upper theta, the value of two A lower value is negative pi divided.
There is also a d r.
We're beating them.
where r is the value zero.
Here comes gradually to two pi.
Get this integral from the first term As you can see r squared divided by two is coming.
I also have two pins at startup.
The dilemma is taking r squared divided by two by two.
Pin the square of a two pi.
The lower limit gives zero.
See there r squared minus here, it is divided by three gives the cube.
There are also a denominator.
When we arrange them We find here again the value.
Look, but do we find a slightly different way.
See 12minus four is eight.
There are also three in the denominator.
PI has a squared divided by the cube.
See the arrival of a square different way.
Above this term has a cube.
There are a follows.
Simplify each other that they Coming to a square.
Whereas in the previous case, a square coming in directly.
Two integral to the same we see that at ease.
To think about it, maybe a little may be more complicated.
But we arrive at the same conclusion, as it should be.
Because the integral sum We know that.
This collection, in the direction of a coordinate ago then in the other direction if you make or
in other if you make You can find the same result.
A collection of how the number of If you collect the same balls
conclusion of to the integral image.
I'm giving you an example homework.
Of the previous example embodiment.
Here's the spiral helix drop
The moment you find the spiral I want two way.
With the first integral by r with integrally by the theta.
Do not make an account here There are cases where it is difficult.
But the important thing here integrals can not find, it is important to write.
Already crooked as an integral Although we can not account closed.
We can do on the computer.
But they do this integral I need to write correctly.
This is an example to show them to you.
Some of the integral may not be easy.
But they list a variety of integration advantage you can find.
Number of numerical results If you're wondering.
A rose curve.
This is an integral, this rose curve find the area of ??a leaf.
It is here, on the with integral will do.
Because for the first integral on the theta
integrally harder I will show that it is.
This is how things function?
When theta is equal to zero is equal to zero.
So here we go.
Where theta is equal to zero.
Are increasing theta.
See this at any point To measure the angle to the center of the tie.
When you get to 45degrees two times the sine of 45degrees.
So the sine of 90degrees Or sinus pi divided by two,
because it's 45degrees pi divided by four, a sine pi divided by two.
So here is the value of a find.
If we continue getting smaller.
Indeed, if theta equals pi divided by two, this
When we come to see that again See angled point pi divided into two
When we arrived at the two-theta function Because sinus becomes p.
E, ie the sine pi,180 the sine of zero degrees.
So it is happening again zero.
So knowing the following three points we can draw roughly.
Aa starting from scratch, going After coming back to zero.
Now in these branches, the leaves of this rose in the Let r integral on before.
Here boundaries enable us important.
is meant to take on the first integral, means to secure theta.
Theta is fixed when we made from scratch comes up to the value on the curve.
is zero, but given the value on the curve.
R This function in terms of theta in spherical coordinate value.
As you can see here will be r squared divided by two.
Here is a place and sinus Put a square of two-theta,
but also b r squared divided by two from two There denominator, a square divided by two,
There are also a sine squared two theta.
This is also here as an integral structure, E value, the value of pi is coming.
That is why this value to us, gives the result.
Now before we do it on theta what if we had to integrals
I suppose it would be.
Refer to this function To find theta aa
We need to partition the set r .Here sinus had two theta.
Inverse sine function be here when you put the two-theta.
It is also divided by the arc sine of a data.
Tetay also divided into two by two,
I'm here to give you the results as is divided into two arc sine becomes divided.
Now if we do on theta We will keep a constant r here before.
Wherein the mean value where the value of the future.
Here are symmetrical.
Less than arc sine arc sine plus the future.
Limits like that.
He grieves, Follow any r de r was here.
is to start from scratch as the smallest, RA will be here at the time.
As you can see is, a reset values Among reset the sine of pi divided
When it comes to half the value of r, it also allows us to arc sine.
This is among the variables that We want to write integration.
Yet a subsequent assignments.
Yet you do this integral In the previous example in the homework
I can not wait here.
But you can write the integral of paramount importance.
Both are developing the ability to think.
What kind of both integrals involved?
This is an important thing.
Integral to calculate separate as a business, and this course
Not too much of goals.
We want to calculate the area of ??the heart curve.
Heart curve, see teta'y once was a minus cosine.
Again, let's try to figure this curve.
Theta zero, the cosine theta one.
So a minus zero.
Here that theta is equal to zero, 're starting from scratch to run.
Theta is equal to the pie, When we arrived pi divided into two,
i.e. when it is on the y-axis of the cosine p divided by two, the cosine of 90degrees is zero.
So it falls to the cosine of theta, just a staying back.
When we came here we go.
So when we came here we go theta equals pi is happening.
So,180degrees.
If we consider the circle at the eastern end of theta zero.
At the western end of theta equals pi.
There is also a negative cosine theta.
So a minus, minus two is happening here.
So you two are going on here.
And this function by theta is symmetric in the six
it becomes symmetrical.
Such a curve is involved.
Already why they call it the heart curve hence this is the structure.
Now let's write the limits of it.
Is still on the integral It will be easier.
Because here straight There trigonometry functions.
If we tried to run the calculation in terms of theta will be made, but the arc cosine.
More will also be another term.
The term for it would be more complicated.
His is the first integral on It will be easier to do.
r limit begins from here.
Keep the curve of theta Coming up on the spot.
r is at zero on the curve kosünüs a time value minus theta.
R d r still there.
That's when I made an integral r squared divided by two to come.
This two in the denominator.
Denominator contains a square here and in the future will be a minus cosine squared theta.
That's got to do this integral.
It's not hard.
If you open a minus two times the cosine of theta, square cosine theta.
A full cosine theta in acres of zero data.
Because the negative with positive values values ??cancel each other.
Or it reverse numerical If you let the cosine
sine of theta theta integral.
Sine of theta zero in two pie.
Zero at zero.
So zero minus zero is going on here.
Zero.
Square cosine of theta angle in the gene pair When using the formula of one-half to come.
The cosine of a future two-theta.
Two cosine theta will fall.
Because I still receive a full acre.
Than one-half of the future p.
Because you will put two pins.
When we account them pi divided by two to three turns.
As I told you on the first theta
get quite complicated integral will provide integral.
This integral is again emphasizes the importance of the order.
In practical terms, or other you can do the integral,
If you do you will find the same result.
Yet, rather than where the account of the integral who want to have a written assignment.
Several moments here Waiting for your writing.
If you are wondering so what would be the values??,
the values ??given here.
But the main objective here is that To write the integral.
Now here is to again, unless there is a slightly different geometry.
An outer fixed e, r is constant curve that has a circle.
It's been two radius.
The inner curve of the heart curve a'yl here.
Because I saw that theta is equal to r pide arrives at a value of two.
As you can see these two tangent.
Now we need to write these limits.
Is still on the integration easier.
Here again because of the inverse cosine, arc cosine was next.
But it's important to limit the summer.
We take a fixed value of theta plotter.
This is a hard right when theta.
The value of theta from the heart line on a circle of radius two.
See on the heart line of the two r d is the value on a circle.
If theta is starting from scratch, coming here and coming up two pie.
For a complete turning 're getting around the center.
This value is put into place, the work E, r r squared integral divided by two.
The upper limit, lower limit.
Put them in place, that the four minus one minus cosine squared theta.
hence comes a square.
There are two in the denominator.
To calculate the remaining integral If we open it to these values ??at work
and here we arrive at the five pi Once a frame is divided into two turns.
It's just easier to provide.
Because these two curves We know the space between.
The outer circle of the inner heart of the area
Remove the space if the curve We need to find these results.
Indeed, the outer circumferential area the area confined by the outer circumferential
pi squared minus four heart the area of the curve limit.
His three pi divided by two We knew it was a square.
The difference between these two really this directly gives the value we found.
Now I have seen quite a few integrals.
The aim here limits When the variables are previously
onto or from the integral over theta which would be easier and
thinking in a circular coordinates,
gücünü to consolidate our thoughts, has to increase.
I think we've come a long way here.
These examples are capable, who understand will have come to a very important place.
I leave here our lesson today.
After that bit different We will now practice.
These applications will be related to derivatives.
Some topics in derivatives, as you can see As three sections, we will see them.
Bye for now.
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