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 II: Karşılaştırmalı koordinat uygulamaları
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Del curso dictado por Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
9 calificaciones
Koç University
9 calificaciones
De la lección
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.
Conoce a los instructores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Prior to the Cartesian coordinates have made examples.
Cartesian coordinates why we use shape
that of the boundary planes.
Here are the spheres.
Of course, the most natural coordinates spheres It looks spherical coordinates.
But with this cylinder coordinate We can see the work can be done.
Of course we know the volume of the sphere.
But this cylinder coordinates, in spherical coordinates and rotating objects
with the quality of these formulas to account We see how to use it.
We'll see.
Larger in spherical coordinates r will go up to scratch.
The angle theta is the latitude bo
ml to scan from scratch two pie will go up.
We have this fi belirliyo longitude angle.
This pie will go up from scratch.
Cylindrical coordinates the Let b as the cylinder.
The radius of this cylinder variables within the sphere.
However, scratch here I was going to a.
D linked, depending on z.
z from the equatorial plane where the horizontal r from the plane arrive.
But below water, too.
This is the minus sign.
So z'yl ties are linked.
There are still zero theta two pie will turn up.
Now let's apply them.
I want to do them in a row Let's see what the differences and
there are similarities.
Sphere volume in rectangular coordinates,
x squared plus y squared plus the surface of the sphere z We know that the square is equal to a square.
In cylinder coordinates is x squared plus y squared
r square is square plus z squared is equal to a square.
The situation in the global koordinar very simple.
R is the surface of a sphere.
As surfaces of revolution If you already see here,
If you r account here will be a function of z or
If you resolve to z r as a function of will.
So this rotational surface In the future into account.
Now let's write in spherical coordinates.
Volume element in spherical coordinates f d r d r squared sine theta d was fi'y.
Now you have two options here.
Before we can separate these theta.
Then fixed limits.
r equals zero ae, theta equals two pie from scratch.
Theta equal, f is equal to zero as pie.
Therefore, these integrals that seems to be so divided.
Because the limits fixed.
When it comes to cylinder coordinates theta and r on the hard limits, even though
See if located in our thought In our bottom-up this coming.
E cylinder coordinates Karey has a squared minus z r.
Below lower halves r squared minus minus sphere
r squared plus a squared minus the square above.
This zoom in on the integrals.
On the reverse before it is We were also able to integrals.
He is already on the surface of revolution speaks for itself.
Above two pi times the rotational surface surface was negative following surface.
R squared minus a squared above the surface,
The following is a minus for this comes twice.
or Z is on r'yl times d integrals squared minus z squared times pi.
Because r squared here.
Diameter, radius of the square there.
z is equal to a negative way from.
Now let's account of how these integrals?
Limits in spherical coordinates can be allocated to the fixed and function.
Our fonksiynu r squared sine fi.
Independent of the sine f r, is independent of the others.
Therefore, we distinguish sinus plug, We are dedicating the square, is dedicating to theta.
d theta pi comes from the two.
r comes from the square cube is divided by three.
ade and new accounts will set r.
The following does not contribute to the b.
Divided by a cube comes from the top three.
Here's a sinus plug minus cosine integral fi.
Minus the value of the cosine of pie.
B minus in there.
There was less than plus or minus one.
B, plus there is coming.
Two.
This integration gives the two.
We know pi divided by three to four times a cube.
Cylindrical coordinates are as follows:Tete simple integration on anyway.
He is leaving.
From here, minus a squared minus z We take r squared plus.
The term comes twice.
z, z, because one of the integral.
One of the integral in the z.
The upper value minus the value at the bottom.
Then we'll hit r d r'yl.
How do we do it?
This is a common type of integration.
If we say that u r squared minus a squared d r d r u would be minus two times.
See here there are two.
There r d r.
B minus the difference here with There are other means.
This is so integral We are putting a minus here.
to force a split second.
Where r is the r by two, giving other.
If we take this integral has minus two pi.
This is minus the return to plus easily be seen.
one-half of the integral.
The top will force an attached.
Two by two thirds to three-box, divided by three slashes comes in two divided by a factor of two to three.
Of course, when R is A,
u is when R is A As you can see above zero.
R is zero below when it is equal to a square.
From the above values ??to zero before Coming to contribute zero
When we gave the following values ??minus.
BI also have negative here.
Plus it was.
See here four thirds pi occurred.
the square root of a.
But we'll take the cube of b.
At that time giving a cube here.
We arrive at the formula we know.
This circular cylinder using the coordinates.
There is a way for sale.
Here are first integral on to be considered.
We know that z squared plus r squared equals a square.
R means, be the square root of a squared minus z squared.
These values ??are up to scratch.
So this cylinder The second way in coordinates.
After picking up our first integral is over
The integral over here before r on the z on.
See where r is the integral square.
Is a square divided by two times.
When you give the square root of r is to remove the square root.
will be a squared minus z squared.
give zero when r b do not contribute to the bottom.
See this a squared minus z is multiplied by the square z'yl.
z goes from A to A minus.
Here we consider the sphere z goes from A to A minus.
As you can bury the dilemma here is cancel each other out a pin remains.
z squared times the integral of a square.
the integral of the square of z in the z-cube divided by three.
Less than a to a.
This is twice the single values??, Jesus in the way, the future of the minus sign
When a sheep instead of twice to zoom minus one-third of a cube a cube.
From here, a cube is divided by three to two,
wherein p is two genes for We're known results.
Already this as the rotary bodies work if we had the above functions, it
the following functions or In the second type of integrals over z
wherein the integral of the coming to end.
So here it intervening in rotating objects integrally automatically formula
involved for him.
We can see here the result.
What if he thought it was the easiest, When we look at the situation in spherical coordinates
the integration variables allocated to was transformed into spherical coordinates.
Thus, here we have found the easiest.
No true integration You do not need to do honestly.
Limits on fixed and functions r squared times the sine of the times
As for variables can be removed triplex
integration of three single ply integrally the anchor could be obtained.
There are examples of the second b.
Before that little I want to take a break.
How to do this, the volume of the sphere cylindrical coordinates or
in spherical coordinates or rotary coordinates a bit to digest
get the opportunity to think.
Bi goodbye until more opinions.
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