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 I: Karşılaştırmalı koordinat uygulamaları
Loading...
來自 Koç University 的課程
多变量微积分 II：应用
9 個評分
從本節課中
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.
與講師見面
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Prior to infinitesimal We have calculated the volumes.
In Cartesian coordinates, cylindrical coordinates,
In spherical coordinates, a special in rotating objects such as bodies.
Now here we will see a number of examples.
They are with us which coordinate system our work would be more appropriate,
integrals in which we do would be more appropriate to show.
I hope that the concepts We will assimilate better.
Of course Swatches than simple will be taken to complex, but the
even very complicated complex because I do not want to take in nature,
that there are so many complicated things in technology When are you going to numerical calculations.
Of course I got to know the basic concepts.
But it can be done manually at because this is an extremely important source,
concepts that can teach you the best ways.
Now the first embodiment, There are a function of the simplest.
w function.
x's, y's, in this way given a function of z.
E x squared minus y sinus and he also limits our
You will recall Cartesian coordinates with constant values
x, y and z direction with constant values of the region began.
It describes the prism.
x is equal to one minus a line
but a point on a line in the plane, a vertical plane in space.
x is equal to a plane passing one.
x is equal to minus one in a plane again.
y plane ie those in the four one vertical and two horizontal
six including plane formed by the intersection of a six-sided prism.
Our problem is the following function integral in this area.
What is the meaning of this?
E.g. distribution, of course, where simply selected a function.
In this way, given that this is a fard please a density, increasingly toward the seabed
Or dwindling skyward y with a density like that.
A thermal energy distribution or providing an electric charge distribution
Or a probability distribution Think of it as a function of that.
That on this volume We want to find the total.
Of course, this integral as follows: We are writing function f x y z
d x d y d z'yl will hit.
Limits on the x fixed, hard on y, z fixed on.
This is just the most simple type We see that here.
Fixed limits.
Moreover function had also a function of x, y has a function,
can be calculated as a function of z.
Thus, this three-integral of integral one-ply,
Multiplication of three integral structure can bring.
Because there's no limit what so
function itself is therefore variables do not affect each other.
X is the integral of the square of course.
is integral to the minus y.
z is the sine integral to pardon them.
Limits given here.
Hit them like this We are easily obtained.
Now the second function again the same as the previous bi.
However, this time limit
constant values ??in the x direction and y direction , while the one constant in the z direction
Even if z eÅittirsıf horizontal plane, coordinate plane x minus top
y plus z variables such as A surface definition.
So we d.times.d this function y We will do our integral is over.
But all of them here, of course, where x y z
which is linear for What if we could of.
But if the variable z zero where z is the natural choice as the right
here we find z zero four minus x plus y to the integral.
This will do on our first integral.
Then y or x on the matter, the will do.
Functions can be allocated to variables.
But with the constant value limits is not defined,
that all previous limits components, such as taking
taking the integral of three single-storey bi We can become multiplied.
Let's do this.
The first step, as you can see here We will do our first integral on.
Our function was x squared y z.
here x and y will be affected.
We have only to him.
z z integral square divided by two.
You're writing here.
This z is zero and four negative in the range of x plus y.
Therefore, single storey integral gone.
Just had a two-storey integral.
I'm bringing this y here.
When this function calculator,
z is equal to zero following a course contribution will not hit above y.
The above frame is coming.
Here are a divided by two.
We got out at one-half.
Now we need to do this integral.
This two-storey integral now you know.
In this two-tiered integrally y boundaries of a merger between minus.
Then it will do the integration.
This merger is a minus here Because of little
To simplify calculations I had to choose such.
y is one function decrease.
Therefore, separate them you do not have to account.
Here our remaining term was: Four minus x plus y squared.
X, Y are each independently of terms We believe it to separate, its square,
plus y squared, multiplied by that year, plus four minus two times x times y.
But in a more y'yl multiplied.
y squared.
See terms here y are the only forces.
Therefore, the only forces y y suddenly drops to minus entrgral,
All that remains is the integral term remains.
This term is an integral over the years means that x is hard to say.
Where b x is the square outside.
Here are four minus x.
Two here.
It took one-half of the two.
y the y integral of the square cube divided by three.
And that a negative value will account for.
y is equal to minus one and two
time will change sign two times lower for the future.
Two divided by three is going on here.
Wherein Y is given instead where x is the square for.
I have four.
There are minus x.
The four x squared minus x cube is happening.
The reset value of x.
That the integration of this single-storey, I would easily account.
Two-thirds here.
Four x cube minus x divided by three to four divided by four.
x is zero and the values ??of a.
Results so simply involved.
This is the third example of a homework as is given to you.
There are still zero on x.
on y zero.
zero on z.
But I take the plane.
This plane is equal to x than z equals one, the plane y equals one.
This plane has a volume formed.
We want to have this volume.
In addition, this volume is with v, z, hes We want to calculate the integral.
It has a meaning.
If you divide the integral take it to v
that the weight on the z axis You can find the location of the center.
If you do not want to deal with it you an example of integration.
Now how do we do it?
Now see here a bit more complicated than the previous.
Because z, z on the start of integrals.
Scratch that you find here is equal to z integration leading to a minus x minus y.
But then the remaining integral over x and y.
But this plane, x and y axis z is zero at x
cuts along the line y equals a plus.
To coordinate plane also is There are currently integral on the triangle.
If we write this more evident this figure
We need to do the integral over.
That accounts integrally You are welcome to.
Here we make it easy as shown in these results.
It answers You can control.
The next sample spherical coordinate problem.
The volume of the sphere.
We know all of us.
Four pi squared divided by three.
If a radius.
We want to see how it found.
In this cylindrical coordinates, We want to make in spherical coordinates.
Moreover globe, is a rotating objects.
You receive a circle.
You can returns.
Turns out the surface of the sphere.
And when we do it for We want to find the volume of the sphere.
We want to make this example.
Now I want to take a break.
After which the gene We will continue to sample.
Ball and roller considerably coordinates example will do.
The reason is clear.
Because these shapes in nature and technology very frequently encountered volumes.
Goodbye.
