Line integral around a circle centered at the origin

[rashida564]

Homework Statement

The picture below shows the text

Relevant Equations

Line integral

Hi everyone, I am confused in this question. First I solved it by noticing that the gradient of the function will be zero (without substitution the hit) I got that it's a conservative field so the integral should be zero since it's closed path. Then I solved it by the hit and convert it as any line integral like any normal integral in terms of the parameter which is theta in this case and evaluate it from zero to 2pi. I got an answer of 2pi. Which method is correct and why.

 

[Orodruin]

What gradient? You do not only have one function.

Please show your work and do not only describe it in words.

 

[rashida564]

 

[Orodruin]

Please type out your work. Posting pictures makes it much harder to help you as they are more difficult to read and cannot be quoted selectively.

For what region is your computation of the divergence (note: not the gradient! Gradients act on scalar fields and yield a vector, divergence acts on a vector field and returns a scslar) valid?

 

[vela]

For what region is your computation of the divergence (note: not the gradient! Gradients act on scalar fields and yield a vector, divergence acts on a vector field and returns a scalar) valid?

The OP calculated the curl, not the divergence, but the same question applies.

 

[Orodruin]

The OP calculated the curl, not the divergence, but the same question applies.

This is true and false at the same time. Depending on if you see this as a three-dimensional problem restricted to two dimensions or a pure two-dimensional problem. The three-dimensional curl theorem is a direct consequence of the two-dimensional divergence theorem (also known as Green’s formula). Either way, it is not a gradient.

Edit: Ultimately, this boils down to the two-dimensional Stokes’ theorem for integration of differential forms. In three dimensions, the curl theorem boils down to integrating a 1-form over a closed curve or a 2-form over a surface while the divergence theorem relates integrating a 2-form over a closed surface to a 3-form over a volume. Both of these may be seen as corresponding to the 2D relation between the integral over a closed curve of a 1-form to the integral of a 2-form over a surface. However, it has more similarity to the divergence theorem as the space of 2-forms in 2D is one-dimensional. However, a deeper discussion of this is going to fly way above the head of the OP at this time.

 

[mathwonk]

You are wrong in thinking that a one form, i.e. something like f(x,y)dx + g(x,y)dy), or equivalently a vector field with entries (f,g), must have integral zero around every closed curve whenever d of it is zero, i.e. whenever ∂g/∂x - ∂f/∂y = 0. what is true is that if there is another function h such that f = ∂h/∂x and g = ∂h/∂y, then the integral is zero around every closed curve. This condition implies, but is not implied by, the weaker condition that ∂g/∂x = ∂f/∂y.

However there is a circumstance under which the two conditions are equivalent, namely they are equivalent in any region which, whenever it contains a closed curve, it also contains all points inside that closed curve. (As Orodruin suggested, the proof is to apply Green's theorem in that case). In your example, the functions f and g have denominator x^2+y^2, which means the region in which they are smooth does not contain the origin (0,0). Hence even though ∂f/∂y = ∂g/∂x, there is no guarantee the integral will equal zero around a closed curve that has (0,0) in its interior. Indeed getting a non zero result for the integral of your vector field around a given closed curve thus proves that the origin is inside that closed curve, so you have proved that the origin is inside the closed curve you integrated over.

This phenomenon is a powerful tool in topology for detecting the presence of "holes" in a region, and is the basis for the theory of "cohomology".

 

[Orodruin]

In your example, the functions f and g have denominator x^2+y^2, whichn means the region in which they are smooth does not contain the origin (0,0). Hence even though ∂f/∂y = ∂g/∂x, there is no guarantee the inetegral will equal zero around a closed curve that has (0,0) in its interior.

I was trying to get the OP to realize this - as is the usual modus operandi in the homework forums.

There is also an additional way of interpreting Stokes’ theorem in such a way that it does hold for this case as well, namely I am terms of distributions rather than smooth functions. The exterior derivative of the one-form in this case is then proportional to the delta distribution at the origin. This view is particularly useful in physics modelling.

 

[mathwonk]

oops, sorry, i did not realize i was in the homework forum.