Green's Theorem for a Circle with a Vector Field

[Nick89]

Homework Statement

For a > 0, let C_a be the circle x^2 + y^2 = a^2 (counter-clockwise orientation). Let \textbf{F} : R^2 \ {0} \rightarrow R^2 be the following vectorfield:
\textbf{F}\left(x,y\right) = F_1\left(x,y\right)\textbf{i} + F_2\left(x,y\right)\textbf{j}

Also given:
\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = \frac{1}{\sqrt{x^2+y^2}}
\oint_{C_1} \textbf{F} \cdot d \textbf{r} = 1

Determine:
\oint_{C_a} \textbf{F} \cdot d \textbf{r}
for arbitrary a > 0.

Homework Equations

Green's theorem:
\oint_{C} \textbf{F} \cdot d \textbf{r} = \iint_R \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \, dA

The Attempt at a Solution

It is obvious that we should use Green's theorem, even if it's not explicitly mentioned in the question, but I fear that I'm using it where it is not valid...

Using Green's theorem directly I calculate:
(R is the interior (surface) of the circle C_a)
\oint_{C_a} \textbf{F} \cdot d \textbf{r} = \iint_R \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \, dA = \iint_R \frac{dA}{\sqrt{x^2+y^2}}
= \iint_R \frac{dA}{a} = \frac{1}{a} \times \text{surface of R} = \pi a

This answer is wrong, and my question is actually why?
I don't need the actual answer to the question (I have it right here in fact) but I need to know why I cannot use green's theorem like this.

I can see two possible reasons:
1. F needs to be smooth (0 is not included in the domain of F)
2. The \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} part needs to be smooth (it's now undefined at 0)

Which is the right reason? Or are they equivalent? I can't remember my teacher telling us F needs to be smooth but I expect he simply forgot...

 

[gabbagabbahey]

Hmmmm... does \frac{1}{\sqrt{x^2+y^2}} Really equal \frac{1}{a} everywhere in your region...or just on the boundary of the region?

 

[Nick89]

Oh wow, that was probably the worst mistake I ever made LOL!

Thanks for spotting that... :p

 

[mhill]

from the definition of the problem and since the line integral would not be defined at r=0 my idea is that the line integral is not 0 but 2\pi

it is a similar problem to 'Cauchy integral formula' on the complex plane but know we miss the ' i'

 

[Nick89]

Actually the answer is 1 + 2 pi (a - 1)

Let D be the region enclosed by the curves C_1 and C_a.
For a < 1 we have:
\oint_{C_1} \textbf{F} \cdot \textbf{dr} - \oint_{C_a} \textbf{F} \cdot \textbf{dr} = \iint_D \frac{1}{\sqrt{x^2+y^2}}\,dx\,dy = \int_0^{2\pi} \int_a^1 dr\,d\theta = 2\pi \left(1 - a\right)
And since the first integral on the left hand side is 1 (see problem statement) we have:
\oint_{C_a} \textbf{F} \cdot \textbf{dr} = 1 - 2\pi(1-a) = 1 + 2\pi (a - 1)

And a similar argument for a > 1 yields the same value.