When the integral of total derivative[tex] \oint d U = 0 [/tex]?,

[alejandrito29]

when the integral of total derivative\oint d U = 0?,

when the integral of total derivative
\oint d U = 0?, and ¿why is zero?

and,

when

\oint d U \neq 0 ??

 

[Bacle]

Alex: are you integrating the identity? And, what is the region over which you are
integrating?

 

[rude man]

The closed line integral dU = 0 if dU is an exact differential. U is then a state function and by definition if you end up where you started you get zero.

Think of a closed line integral of gravitational or electrostatic force.

 

[Bacle]

An example , a classical one, I think, of a form that is not exact is that of w=dz/z

defined on the circle |z|=1, embedded in R-{(0,0)}, or C-{(0,0)} .

 

[alejandrito29]

Alex: are you integrating the identity? And, what is the region over which you are
integrating?

my problem is

\oint d( \frac{dA}{dy} exp(A) )

where A=|y|
but, y is a angular coordinate...between )-\infty, \infty(
but my problem is that
\frac{d|y|}{dy} = -1 , y \in )-\pi,0(, )\pi,2\pi(, etc
\frac{d|y|}{dy} = 1 , y \in )0,\pi(, )2\pi,3\pi(, etc
\frac{d|y|}{dy} = undefinided , y =-\pi,0,\pi...k \cdot \pi
........................
but, if i integrate

\oint exp(|y|) ¿why is it Non zero, if i too started and end in the same point (0 and 2pi por example)

 

[Bacle]

Alex:
By a theorem; I think Stokes or one of its corollaries, your last integral equals zero iff , f (exp|y| , in our case) as a differential form, is exact, or, equivalently, if there is an F
in the region of definition, with dF= exp|y|.

I'm sorry, I still don't fully understand why y is positive in some regions, and negative in others. Do you have an explicit formula for it?

Still, if you know that your y is defined as you said, the only reason I can see for why
it is not zero, is that , it does not have a global antiderivative.

 

[alejandrito29]

question :

can i to say d \theta = \frac{1}{x^2+y^2} (-y dx +x dy )
since i am to integrate on a circle, and , since

x = 1 \cdot \cos (\theta) ; y = 1 \cdot \sin (\theta)

and then

\oint e^{|\theta|} d \theta = \oint e^{| \sin^{-1} y|} \frac{1}{x^2+y^2} (-y dx +x dy ) \neq 0 , since is not globally defined in x, y = 0,0 ?

 

[Bacle]

The standard way (the one I know : ) ) of doing line integrals/ contour integration is
by parametrizing the contour:

x=cosθ , so that dx=cosθdθ

y=sinθ , so that dy=sinθdθ

And then integrate from 0 to 2∏ ; since the singularity happens at (0,0), which is
not in the contour, you don't need to worry about this. Still, another way of testing
whether:

dθ:= (xdy -ydy)/(x²+y²)

is exact, is by integrating around a close contour. If the form is exact, it would then
integrate to 0, by an extension of the fundamental theorem of calculus.

 

[alejandrito29]

firts , sorry by many questions,

in my case, i need to integrate on te circle

but in a text says:

(xdy -ydy)/(x^ 2+y^ 2) it is not exact, since its integral along the unit circle is not 0... \int ^\pi_{-\pi} d\theta = 2 \pi...
this is a argument for says that \oint e^{|\theta|} d\theta is not zero??