Understanding Line Integrals with Respect to Delta X and Delta Y

[Cyrus]

Thanks Krab, is there any error in post #57 where I discussed green's theorem? As for the applications, is it merely a tool for simplifying line integrals. Or is there other applications?

 

[Cyrus]

Yet another question I am afraid. I am reading one part of my book and it discusses Vector Forms of Green's Theorem. It shows the curl F dot k is equivalent to greens theorem. Thats fine, but then it has a formula of greens theorem in terms of the normal and divergence. Normally, Greens theorem deals with F dot T, the unit tangent, so it represents the component of force along the directon of moment at all points, and sums it up. But this second form of greens theorem implies that you compute the sum of all normal components and sum it along the curve. I have never seen this before, and don't know of any practical use.

Just a guess, if you want to find the amount of friction done along a path that veries, with varying frictional force values, just integrate, and multiply by the mass to get the total work done?? Not a very good guess though.

One final thing, the book states:

"T(t) = \frac{x'(t)}{|r'(t)|}i + \frac{y'(t)}{|r'(t)|}j

You can verify that the outward unit normal vector to C is given by

n(t)= \frac{y'(t)}{|r'(t)|}i + \frac{-x'(t)}{|r'(t)|}j
"

The dot product is indeed zero, so it is normal. How did they know to put the negative sign in the j component and not the i component? Also, how do they know that placing the negative sign here, apposed to the i component produces an OUTWARD unit normal as opposed to an inward unit normal?

 

[reilly]

.

One final thing, the book states:

"T(t) = \frac{x'(t)}{|r'(t)|}i + \frac{y'(t)}{|r'(t)|}j

You can verify that the outward unit normal vector to C is given by

n(t)= \frac{y'(t)}{|r'(t)|}i + \frac{-x'(t)}{|r'(t)|}j
"

The dot product is indeed zero, so it is normal. How did they know to put the negative sign in the j component and not the i component? Also, how do they know that placing the negative sign here, apposed to the i component produces an OUTWARD unit normal as opposed to an inward unit normal?[/QUOTE]
**********************************************

All you need to do is draw a picture.
Regards,
Reilly Atkinson

 

[Cyrus]

Im sorry but I don't see what you mean. x'(t) and y'(t) could assume positive or negative values, so how could I draw a picture?

Ah yes, I see what you mean clearly now. The thing to remember is that the tangent vector is alway along the directon so that the curve lies to its left. (positively oriented.) This means that it can never have a negative in the i term, it can only be in the j term.

I still don't understand greens theorem for the normal component. I understand math wise, you sum the normal at all points along the curve, but in physics what does this mean!?

 

[Cyrus]

Another thing I noticed was that in my physics book they use the notation \oint when writting Gauss law. But isint that symbol for the integral along a closed curve. In my math book, they use a double integral to represent the flux. How come the physics book does not have a double integral over the region? Which representation would be correct?

P.S. If you could explain an application for green's theorem for the normal component, I would appreciate it.

 

[Galileo]

Oh yeah, and one other thing. I would assume that Greens theorem is only good for R^2? The reason being that it only involves vector fields that have an x,y component. What are some applications for this theorem? My first guess, seeing a line integral around a closed loop would be faradays law. But with faradays law, don't we have to associate a "sheet" when we do the integral, (that trick maxwell played when the problem with flux for a capacitor arose and test charge buisness). I don't see any sheet in Greens theorem except maybe for the one on the planar region.

Yes, Greens theorem is stated explicitly for the plane. C has to be a plane curve and D is thus a planar region.

However, I`m sure you'll get to Stokes theorem soon enough. It can be regarded as a higher dimensional version of Green's theorem. In three dimensions, you have many ways of choosing your 'sheet'. In contrast, in two dimensions D is uniquely defined by the curve C.

Are you using Stewart's 'Calculus. Concepts and contexts'? Your quotations from the book sound awfully familiar.

How come they define the differential operator with the i,j,k on the left hand side of the partial derivative fraction? Then they do the gradient of f, and they seem to move all the i,j,k's from the left side to the right side? Also, the book also uses the differnetail operator with i,j,k on the right side. Whats going on with this inconsistancy? Is it just carelessness?

Actually, it's carefulness.
It doesn't matter where you put the unit vectors i,j and k right?
However the notation:

\nabla = \frac{\partial}{\partial x}\vec i+\frac{\partial}{\partial y}\vec j+\frac{\partial}{\partial z}\vec k

may lead to confusion and to thinking a big error; namely that you'd have to take \frac{\partial \vec i}{\partial x} and such.
In \nabla f this confusion doesn't arrive, so they write it back on the right side again, since that is standard.

Another thing I noticed was that in my physics book they use the notation when writting Gauss law. But isint that symbol for the integral along a closed curve. In my math book, they use a double integral to represent the flux. How come the physics book does not have a double integral over the region? Which representation would be correct?

I've seen that being used for both closed loops and closed surfaces.
It's just that physicists are generally too lazy to write three integral signs for a volume integral and two integrals signs for a surface integral. :zzz:
No confusion will arise if you know what you're dealing with.

P.S. If you could explain an application for green's theorem for the normal component, I would appreciate it.

I've never seen a physical application. But it's useful for proving some identities like:

\int \limits_D \!\!\! \int f \nabla^2 g \, dA=\oint_C f(\nabla g)\cdot \vec n \, ds-\int \limits_D \!\!\! \int \nabla f \, \cdot \, \nabla g \, dA
This is Green's first identity.

 

[Cyrus]

Yes, it is stewart. ;-) However, he pretty much ripped off Swokowski's book Calclus with analytic geometry, because they are nearly identical in text and content. I thought that the greens theorem for a line integral could mean the flux through that line integral. But I don't know of any application of flux through a line. Faraday has to do with the flux through the region contained within the line, not the actual line itself.

 

[Dr Transport]

Greens theorem is used in the Sommerfeld Theory of scattering by an aperature in physicsl optics. In this case g is usually a spherical wave of the form \frac{e^{\pm i\vec{k}\cdot\vec{r}}}{r}

 

[Cyrus]

I gota question on surface integrals. When the book approximates the area of a small patch, it uses the tangent vectors at the edge of the patch. Then it computes the area by using the cross product of these two tangent vectors. It has the parallelogram labeled with each side as \Delta u r^*_u and \Delta v r^*_v. How come they don't use UNIT tangent vectors? If we use a lot of very small patches, then a tangent vector at the edge of each patch may be huge, and when you cross them, you could get a rediculously big area of the parallelogram that approximates the area.