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I need to show that moving the curve to a simply connected region, the integral of the field over the curve will be $0$.

Given $F(x, y) = ((-y)/(x^2+y^2 ), (x/(x^2+y^2 ))$, and $γ$ circle of radius $1$ centered at $(2, 0)$; i.e., $γ ( t ) = (2 + \cos t ,\sin t )$, $0 ≤ t ≤ 2π$. Note that $γ$ is entirely contained in the open half-plane $x > 0$, the half-plane is simply connected, and $F$ is defined everywhere in the half-plane.

How do I show that $$∮F \cdot ds = 2∮_0^{2\pi}(1+2\cos φ)/(5+4\cosφ)\,dφ$$ and subsequently $$∮_0^{\pi}F \cdot ds = π −π = 0 $$ (I know this is true because each path taken is independent)

I tested $∂M/∂y=∂N/∂y$ for conservative field, it checks out. Then I derive the potential functions. However, it seems I need to use the polar coordinates, since I have a circle with points $(3,0), (2,1), (1,0), (2,-1)$ on the $x$-axes and $y$-axes. I just don’t know how to implement what I know.

HELP PLEASE!!

Lilacfluer
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  • See math notation guide. Also, you are giving a list of things to show instead of asking a question. What is unclear to you about these exercises? –  Nov 30 '14 at 05:56
  • I was able to edit the question.. – Lilacfluer Dec 02 '14 at 00:14
  • Okay, I edited it a bit more. So, the first part invites you to plug $\gamma$ into the formula $\int_0^{2\pi} F(\gamma)\cdot \gamma'(\varphi),d\varphi$. After simplification, this should give that integral with $\cos \varphi$. That the integral evaluates to zero follows from the fact that $F$ is conservative in a simply connected domain containing $\gamma$. –  Dec 02 '14 at 00:21
  • Looks so much better, thank you.. – Lilacfluer Dec 02 '14 at 00:24
  • @Lilacfluer Can you find a potential or not? This isn't clear to me from your question. – Git Gud Dec 02 '14 at 00:38
  • I did, $-tan^-1$(x/y) + C...I think – Lilacfluer Dec 02 '14 at 00:44
  • @Lilacfluer It doesn't even matter what it is, all that matters is that it exists. To understand why, see this answer of mine. (It's from complex analysis, but the idea/symbols used on the proof are the same). – Git Gud Dec 02 '14 at 00:47
  • @GitGud I saw it, thanks, but that’s just the general form, which I get...So I’m trying to apply that to the example above, but struggling to apply the general formula – Lilacfluer Dec 02 '14 at 01:17
  • @Lilacfluer Let's see. So you understand that $\displaystyle ∮F \cdot \mathrm ds=0$. You just don't understand why $\displaystyle ∮F \cdot ds = 2\int \limits_0^{2\pi}(1+2\cos φ)/(5+4\cosφ),dφ$, is that it? – Git Gud Dec 02 '14 at 01:18
  • well... I think the circle has to be split in half. I’ll try to work out. – Lilacfluer Dec 02 '14 at 01:21
  • Let us continue this discussion in chat. – Lilacfluer Dec 02 '14 at 02:03
  • @Lilacfluer I can't join you in chat now. Going off. Good luck. – Git Gud Dec 02 '14 at 02:08

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