I'm trying to find the value of $\int_c \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$ along the curve $c(t) = (1+2cost,sint), 0\leq t\leq 2\pi$
I found the potential fuction $\phi = \tan^{-1}(\frac{y}{x})$ $s.t.$ $\phi_x = \frac{-y}{x^2+y^2}$ and $\phi_y = \frac{x}{x^2+y^2}$
Therefore, my answer is $\phi(x,y)]_{c(0)}^{c({2\pi})} = \phi(2,0) - \phi(2,0) = 0$. But the answer I checked was $2\pi$. What did I wrong in my solution?
Regards
Your potential function $\phi(x,y) = \tan^{-1}\left(\frac{y}{x}\right)$ is discontinuous on the line $x=0$, which $c$ crosses twice. This is the reason your method failed.
To redeem it, we may break $c$ up into three parts: (1) $0 \leq t < 2\pi/3$, (2) $2\pi/3 < t < 4\pi/3$, and (3) $4\pi/3 < t \leq 2\pi$, and add up all the results, giving
$$\left(\lim_{x\to 0^+}\phi(x,\sqrt{3}/2) - \phi(2,0)\right) + \left(\lim_{x\to 0^-} \phi(x,-\sqrt{3}/2)-\lim_{x\to 0^-} \phi(x,\sqrt{3}/2)\right)+\left(\phi(2,0)-\lim_{x\to 0^+} \phi(x,-\sqrt{3}/2)\right) = \frac\pi2 + \pi + \frac\pi2 = 2\pi$$
If it helps, your initial attempt is similar to the one-variable error: $$\int_{-1}^1 \frac1x\,dx \color{red}= \ln|x|\Big|_{-1}^1 = \ln|1| - \ln|-1| = 0$$ where again the problem is that the "potential" $\ln|x|$ is discontinuous at $x=0$. The fix here is to split the interval at $x=0$ and evaluate as improper integrals. Your problem has the same fix.