- #1

kostoglotov

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## Homework Statement

Firstly, I was seeking any clarification on whether I've made any mistakes. Secondly, further insight into Green's Theorem, if my working is all good.

I've been reading the mathinsight.org on Subtleties about curl: http://mathinsight.org/curl_subtleties

Regarding the vector field [itex]\vec{F} = \frac{1}{x^2+y^2} \langle -y,x,0 \rangle[/itex] I decided to test Green's Theorem out.

This page talks about how this vector field has macroscopic circulation but no microscopic circulation.

It would seem that for this vector field [itex]\oint_C \vec{F}\cdot d\vec{r} \neq \int\int_D (\nabla \times \vec{F})\cdot \hat{k} \ dA[/itex]

## Homework Equations

## The Attempt at a Solution

So [itex]curl \ \vec{F} = \vec{0} [/itex] so [itex]\int\int_D (\vec{0})\cdot \hat{k} \ dA = 0[/itex] regardless of our path C or domain D.

So let's consider the unit circle as our path

[itex]C: \ x^2+y^2=1[/itex] for [itex]\oint_C \vec{F} \cdot d\vec{r}[/itex]

Standard polar form conversion [itex]\vec{r}(t) = \langle cos(t), sin(t), 0 \rangle[/itex]

So [itex]\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi}\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt = 2\pi[/itex]

So Green's Theorem doesn't work for all vector fields, even if it's a simple enclosed region...is this due to a discontinuity at (x,y) = (0,0)?