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
Çözümlü Problemler III: Karşılaştırmalı koordinat uygulamaları
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Do curso por Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
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Koç University
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Uzayda Yüzey ve Hacım Entegralleri
Uzayda yüzeylerin açık, kapalı ve parametrik fonksiyonlarla gösterilmeleri ve eğrisel koordinatlar. sonsuz küçük yüzey alanları seçeneklerinin birleştirilmiş bir yaklaşımla elde edilmesi. Uzayda küre, koni, paraboloitler gibi temel yüzeylerin tanıtılması ve bunları içeren alanlarla hesaplar.
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi there.
Now before this one We have calculated the volume of the sphere.
That in spherical coordinates and cylindrical coordinates we did.
In addition, a sphere rotating objects We have calculated and varied as
coordinate systems interact 're doing, we see how.
Here sphere similar to the problem, but We're doing a bit of a different problem.
There are still a sphere.
But this sphere with a cone're taking a nap.
So as for zero angle take a right which our
rotate around the axis when 're getting a cone surface.
Now this cone with the surface of the sphere We want to find the rest of the volume.
Of course, on this sphere Or intersecting spheres fi
Putting zero to zero will be zero volume.
Putting PI divided into two semi- will be the volume of the sphere.
This tells us one of our accounts gives you the ability to control.
Fu took the pie get zero,
If we grew up, then it will be the volume of the entire sphere.
As we see in this case, however see:elevational view of this sphere.
This cone to download and view.
F is the zero angle.
Wherein f is the zero angle.
Of course, this is cutting at a height b.
This occurred at the height of b There are terms for zero and
Create it in the long run to zero.
These cylindrical coordinates you'll have when using.
Now that three different genes To calculate the.
Here drawn the same way.
Now and in the sphere and the cone surface and global koordinar
cylindrical coordinates define.
Also as rotating objects,
we need a bottom-limiting or top curve was limiting curve.
Let's move on them to account.
Now in spherical coordinates certain sphere equation.
R is A.
A hard.
It is as simple as going to.
Forever in spherical coordinates We know little volume.
We know that the equation of the cone.
The equation for the cone is equal to zero at f.
As you can see in spherical coordinates again began to fixed limits.
When it comes to cylinder coordinates r squared plus z squared would be a square.
Because these two This is when we combine
squared plus z squared gives a squared.
We have seen it before.
Certain volume in the cylinder coordinates.
We write the equation of this cone need to coordinate the cylinder.
Roller cone coordinates equation a linear equation, z r plane.
If this height, b,
If r is zero in the horizontal, who is possessed with split r b r is zero.
Here r is zero divided by the slope b gives the tangent of the angle here.
This is the work of our describes the relationship.
Now the top and bottom Considering the limits,
As the rotary bodies, lower limit that right.
The upper boundary of this circle.
We found the correct equation here.
The equation of the circle is squared plus z that is equal to a square by square
z equals r squared minus a squared i.e. as a function of r.
This rotary bodies
In our calculations we will use size.
Now we need to find the limits of integration.
Sphere is very simple.
The boundaries are integral.
A scratch is going to.
Fu also going to zero for zero.
In the following we need the cylinder b.
r goes to zero is zero.
But this is our
from the surface of the cone i.e. z r b is equal to zero divide,
the right of this equation, in the circle
to the equation z squared r squared minus goes.
This is the integral over Would it be appropriate to do?
However, because we are here We accept the following.
We are writing the limits of z r.
However, in contrast r z we could write out the boundaries.
But I see here is a limit to scratch is going on this cone.
Here scratch circle can go on.
This can be written, but a You can not get by integrals.
You must write two integral.
This does not seem very appropriate.
The two integrals can do, instead of making a single For that you will choose to do with integrals.
Now in spherical coordinates
like this:Once variables a separable function.
In addition, the fixed limits.
Theta of zero to two pi r Reset between fi reset the zero of f.
Therefore this is the easiest global There seems to be at the coordinates.
Cylindrical coordinates again theta, but are separated from scratch
r is zero and z is from the surface of the cone,
The equation of the circle from the right Coming up to the equation.
So here's one of the z, z on integral, but limits vary.
We'd like to do as solids of revolution
r squared minus a squared circle of the upper limit,
towards lower boundary of the cone, cone produces the correct equation.
b r r is zero divided.
Where f is obvious.
We know the formula.
g is clear.
r r d.
Yet here is the integral over
z to later When you want two
various other ways to be an integral do not get on our integral type.
Here also integral emerges.
Now let's account for this integration.
In the first integral variable was divided into.
This is a two theta pi comes from.
Coming from the square is a cube divided by three.
FI minus sine cosine integral fi.
hence the above minus cosine fi zero.
therefore, the following is in the negative for.
Replace them, we can arrange some more.
There are two pi divided by a cube here.
But this, see reset a minus cosine fi multiplied.
i.e. b d b is the bottom of the cone since it was high.
It's a times cosine
See if we use that for this to be zero Disconnect one of the cube, where a there.
Here a times cosine fi will b zero.
a minus b.
Here, too, will be a square.
There are two pi divided by three.
Or a zero minus cosine for two
times for sine-squared divided by zero two as you can.
Here you have a cube.
We concluded with them.
As we can control:the br br see if zero was this height.
If b is zero it down.
Remove half the volume of the sphere.
b has to be equal to zero.
Because b closer to See This cone is getting tighter,
Meanwhile there will be a volume is shrinking.
also if you get a D minus, See the full sphere that time comes.
Indeed, when we look at here, when b is equal to a zero.
b is zero, when the volume of the hemisphere,
when b is equal to minus Turns out the volume of a full sphere.
In cylinder coordinates the situation is somewhat different.
Refer to the following integral We need to have accounts.
This upper limit.
This lower limit.
This integral is not very difficult.
We have done it before.
If we say that u r squared minus a squared, u would force a split second that.
of r d r a d u'y gives the negative half.
Bi geliyo two factors and the minus sign.
There are two cons.
There are one-half.
This is a split second force the second would be divided three integral force.
But I need to divide the three splits in two.
Thus two thirds coming.
This is easy.
b, and r is zero constant.
Then the r squared are also divided by three integral cubes.
r is zero is true if r is zero for the cube is going on here.
Gene simple geometric properties Using wherein the required
simplification doing just that The same formula is obtained.
As you can see in d If equal is zero.
If b is zero interest hemisphere.
If b is equal to minus the Remove the outcome of a full sphere.
Here is a sample you added I give homework to do.
These examples on your own work pekiÅtirebilme lets you thoroughly.
There is already a difference from the previous one b.
This body weight on the z-axis around the z axis with the center
moment of inertia, d v r squared second moment.
Again, as I always say them in random functions
but b in selected technology,
Examples are similar to those encountered in nature We have received so simple integration.
Our goal is already very complex integrals not calculate these calculations
Specific examples will reinforce the view method, a little bit of your other courses in physics,
technology courses that you'll encounter in Do not integral types of accounts.
No results as the control will also have the opportunity to.
The second problem is similar to the first but with a cone cut
b z is the surface of the sphere cut with a horizontal plane.
So this is not exactly a dome, a partial dome.
It is also used in many places.
You want us to find the volume.
Now it coordinates two types of can be calculated again.
In cylinder coordinates or spherical coordinates.
Let's see which one will come easily?
I can already see This side view.
when z is equal to b cut cylinder
coordinates this The definition of boundaries is very easy.
z is equal to b.
Above the surface of the sphere is an easy to.
But we'll see, but this is a very difficult global
coordinates of this surface Definition little more complicated.
Yet here's the tip geometry.
So if b is zero half-dome.
Half-ku, with sphere dome.
If you see a backwards b is equal to b There is nothing grows grows.
Has to be zero at the end.
If you get a minus instead of a full sphere b I need you can not find the full results.
It looks a bit like the other.
At the other, with a cone cut dome that was generated
If the above plus Kubba following cone.
Here are just above the dome.
How will this summer integrals?
Now spherical surface and the horizontal plane
equation in cylindrical coordinates bit easier.
B z is the horizontal plane equation z is in spherical coordinates r
Once your rib was fine.
According to z b is equal to the global coordinates the equation of the horizontal plane.
As well as in terms of r f You can also account.
The following is the account can be denominated.
But in spherical coordinates There is a convenience.
Supremely simple equation of the surface of the sphere.
Here, a simple horizontal plane.
Here a bit more complicated horizontal plane.
In the global koordinatar sphere equation is simple but
cylindrical coordinates of the sphere equation is not equal to a constant.
Now that ha, questions may come to mind.
Which coordinates are more appropriate here?
Global or cylindrical coordinates for?
As you can see here some a simpler, more difficult here.
Here is more difficult, more simple here.
One of the integral in which to do?
Here you have fun.
Let fi Zera integrated to the first, run on before?
There is here also there z.
Once on z, on or before r?
Therefore, they decided should be given topics.
In the previous problems It was easier to make these decisions.
Now let's do the following here: Our goal is to learn these things.
Let's do so well in both cylinder,
Let's also on the sphere and Let's see how that develops.
If the parties have little Let's see how we handle it.
Now before you roll 'll start by coordinates.
Yet again, I have two questions here.
Let z integrated on or before Let's first integrated run on?
Now that the first step was to write the equations,
in that the boundaries of dome above the plane below.
In the second step, let's write these integrals.
Let's write.
First, let's write on the cylinder.
Here on the cylinder There are some preparations that.
z equals zero zero b in space.
a circle of radius r is zero occurs because
When you cut here b r is zero if this is obvious.
a squared minus b squared.
Now before we get integrals over z, d z will be.
where z is coming from?
i.e. b from the plane of until the surface of the sphere.
You're writing.
Here we look at the upper limit is variable.
And then on the r We're taking the integral.
r r d. Where r is the zero-zero constant
value is going up.
Tete already leaving.
This is the third step in the alternative,
If we look at the option, on the first run Although we have made ??will be integral where r d r.
b is zero as a function of z
see where it will go up to When we get here before we set r
If we keep fixed constant from z comes to the surface of the sphere.
On the contrary, prior to If we take if we keep z fixed,
z scratch the surface of this sphere is coming.
Then from there ae z will go up.
Here, we see them in these applications.
So the third step of this gene as the base for my show
cylinder surface again before integration is done on.
z is from boundaries easyto A constant values ??as you.
Tete is leaving again two pi.
Now let us calculate the integrals.
Now on z, and theta accounts, easy two theta pi.
r is zero to zero.
But see here the top There are negative values ??lower value.
Here is a square minus the square There minus the square root of b.
We stand with r d r.
This type of integration always We encountered a r squared minus
If we say that u r d r squared minus two would be d.
So here is r d r b minus and b d u here with two differences occur.
If we do them properly and If we write the proper limits
here it is a divided Is the second power d.
three divided by two thirds will be divided by two.
We've seen this before.
Here the constant b.
There r d r.
He also had two squared divided by r.
Set limits by going this way.
We arrange them When we come to this conclusion.
Interpret again.
If b is zero, see also b is zero or the means,
dome, intersecting spheres intersecting z is zero to be on the surface,
to be in the horizontal plane, b is zero we obtain the half-sphere.
Its area is also really two pi We know that a cube divided by three.
b goes to zero, b is zero go to b means that the volume is reduced by yükselttik.
As a result, b, b Go to pardon me if
I said it wrong, is going to b When the volume has to be zero.
b, when a given There are two a cube here.
There is a cube here.
Three. A tab
a cube, but comes with a minus sign.
Zero is coming.
Here is a minus b, giving it that way You can also check again.
b if a negative three a cube is happening here.
Less data as one goes from here.
Two stays here.
Two a cube.
Wherein two have a cube.
Known sphere divided by four pi as a result of a cube turns three.
Hesaplasak same thing as rotating objects.
In this cylinder coordinate The final step is happening.
Let's look at one of the global coordinates.
In spherical coordinates theta zero two pie, every time we encounter it.
Let us take first the integral over r squared.
r limit r plane was divided by b is equal to the cosine fi'y.
Because z is equal to r times the cosine fi.
z is equal to B. For plane intersecting the sphere.
We went to the other side of the globe r is equal to the time we arrive.
So is this coming from the integral over.
See, here is a fun.
I also have here a fine.
Now get rid of this integral is also theta.
Where r is the square cube is coming from.
cube is divided by three.
The upper limit.
That's easy.
But following the cosine of the border there.
Cosine cube is coming.
This integral on f not very difficult when you have made.
FI minus sine cosine integral fi.
That's easy.
This second term is not very difficult.
Cos u say because your plug, u the negative third power.
Here the minus sine fi d fi d u will.
So it is integral The integral is minus two.
including the cosine.
That term here.
These terms
When we get organized now We are:a times the cosine fi
v is zero if düÅünüÅün, Let one here here.
Here would be a squared b.
Here twice a cube.
Here would b cube.
Altogether, we found a previously we obtain the formula.
Which is easier?
Maybe it depends on the person.
In cylinder coordinates integration more difficult.
Because under this karekökl ??expression.
Is made a little more terms.
Here it follows the cosine fi in the denominator,
have the lower limit of Business but looks like it will mix,
where cosine cube, as well as a Because it is one for the sinus,
If u u minus the cosine Once there is a third force.
Therefore, it is easy this can also be integrated.
Now here I want to give homework.
If you remember we were on the cylinder is integral over before we did.
From you, on our first what would happen if the integral?
You and your writing and computation course I'm waiting for you to find the same result.
The dome on the sphere,
demi is before we've done on the integral, In the example we have just solved it.
I give you homework, Sign on for the first integral.
It will be easier to see.
It will also be easier to see.
Slightly heavier part As I drove parsed.
If you make them a very will absorb a good subject,
You will learn to write in and the answer is certain to
able to compare with others You can also see and options have been happening.
There is a duty.
This is the same for the following types of dome the integral z who is possessed with the
who is possessed with is the integral of the square.
Their answers are given.
i z gives the following.
Weight on the z axis coordinate of the center.
About an axis that
inertia of a rotating dome showing the moment.
Now I want to take a break again.
We have solved similar problems in succession two.
This cone looks like the problem to a great extent we solve the previous problems.
But I'm a breather.
You will have the opportunity to take a look as well I hope that just the parsed problems.
And in what used to be homework write formulas if at all,
you write the integral of the integral You can see that it is easy.
Goodbye until the next session.
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