(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For a > 0, let [itex]C_a[/itex] be the circle [itex]x^2 + y^2 = a^2[/itex] (counter-clockwise orientation). Let [tex]\textbf{F} : R^2[/tex] \ {0} [tex]\rightarrow R^2[/tex] be the following vectorfield:

[tex]\textbf{F}\left(x,y\right) = F_1\left(x,y\right)\textbf{i} + F_2\left(x,y\right)\textbf{j}[/tex]

Also given:

[tex]\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = \frac{1}{\sqrt{x^2+y^2}}[/tex]

[tex]\oint_{C_1} \textbf{F} \cdot d \textbf{r} = 1[/tex]

Determine:

[tex]\oint_{C_a} \textbf{F} \cdot d \textbf{r}[/tex]

for arbitrary a > 0.

2. Relevant equations

Green's theorem:

[tex]\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[/tex]

3. 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)

[tex]\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}}[/tex]

[tex]= \iint_R \frac{dA}{a} = \frac{1}{a} \times \text{surface of R} = \pi a[/tex]

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 [tex]\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}[/tex] 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...

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# Homework Help: Green's theorem question

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