# When the integral of total derivative$$\oint d U = 0$$?,

1. Sep 23, 2011

### 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

$$\oint d U \neq 0$$ ??

2. Sep 23, 2011

### Bacle

Re: question

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

3. Sep 24, 2011

### rude man

Re: question

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.

4. Sep 25, 2011

### Bacle

Re: question

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)} .

5. Sep 25, 2011

### alejandrito29

Re: question

my problem is

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

where $$A=|y|$$
but, y is a angular coordinate...betwen $$)-\infty, \infty($$
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)

Last edited: Sep 25, 2011
6. Sep 27, 2011

### Bacle

Re: question

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.

7. Oct 1, 2011

### alejandrito29

Re: question

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 $$x, y = 0,0$$ ???????????????????????????

Last edited: Oct 1, 2011
8. Oct 1, 2011

### Bacle

Re: question

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
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.

9. Oct 2, 2011

### alejandrito29

Re: question

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????????????????