Independence of Circular Contour Radius - Part 1 (link to part 2 of question)
Problem:
Let $P$ and $Q$ be smooth functions on $D$ satisfying $\partial P / \partial y = \partial Q / \partial x$. Let $\gamma_0$ and $\gamma_1$ be two closed paths in $D$ such that the straight line segment from $\gamma_0 (t)$ to $\gamma_1 (t)$ lies in $D$ for every parameter value of $t$. Then $\int_{\gamma_0} P$d$x + Q$d$y = \int_{\gamma_1} P$d$x + Q$d$y$. Use this to demonstrate that, by using Green's theorem, $\oint_{|z|=r}P$d$x + Q$d$y$ is independent of the radius $r$, for $a<r<b$.
An attested solution$^1$ to the problem can be found on the Net, which states that—in the words of Gamelin$^2$—provided we let $D$ be a domain, and we let $\gamma_0 (t)$ and $\gamma_1 (t)$, $a \le t \le b$, be two closed paths in $D$, and if we suppose that $\gamma_0$ can be continuously deformed to $\gamma_1$, in the sense that for $0 \le s \le 1$ there are closed paths $\gamma_s (t)$, $a \le t \le b$, such that $\gamma_s (t)$ depends continuously on $s$ and $t$ for $0 \le s \le 1$, $a \le t \le b$, then since
$\hspace{2in} \int_{\gamma_0} P$d$x + Q$d$y = \int_{\gamma_1} P$d$x + Q$d$y$
for any closed differential $P$d$x + Q$d$y$ on $D$ we can use straight lines to deform $\gamma_0$ to $\gamma_1$ having defined $\gamma_s (t) = s \gamma_1 (t) + (1-s) \gamma_0 (t)$, $0 \le s \le 1$, $a_1 \le t \le b_1$, for which the above theorem applies. The solution goes on to say that "by parameterizing the circles $|z| = r_0$ and $|z| = r_1$ by $\gamma_0 (t) = r_0 e^{2 \pi it}$, $\gamma_1 (t) = r_1 e^{2 \pi it}$, $0 \le t \le 1$ [the] straight lines segments joining $\gamma_0 (t)$ to $\gamma_1 (t)$ are radical [and] are in the annulus."
[1] (p.015) http://tinyurl.com/ad2e62c
[2] (p.082) http://tinyurl.com/anmmcq8 
[*] (p.222) http://tinyurl.com/bxhb866
[*] (p.320) http://tinyurl.com/ayhsnko 
 A: The assumption on the force field $K:=(P,Q)$ guarantees that $K$ is locally exact. By Green's theorem this means that for any simply connected$^1$ subdomain $\Omega\subset D$ the integral $\int_\gamma(P\>dx+Q\>dy)$ along any closed curve $\gamma\subset\Omega$ is zero. Write $P\>dx+Q\>dy=:\omega$.
Assume $t'<t''$ are nearby values of the parameter $t$. Then there is a simply connected subdomain $\Omega\subset D$ containing the arcs $$\alpha_0:\ t\mapsto\gamma_0(t),\quad \alpha_1:t\mapsto \gamma_1(t)\qquad (t'\leq t\leq t'')$$ as well as the segments $\sigma'$ and $\sigma''$  connecting $\gamma_0(t')$ with $\gamma_1(t')$ and $\gamma_0(t'')$ with $\gamma_1(t'')$. As $\lambda:=\alpha_0+\sigma''-\alpha_1-\sigma'$ is a closed curve in $\Omega$ it follows that
$$\int_{\alpha_0}\omega -\int_{\alpha_1}\omega=\int_{\sigma'}\omega-\int_{\sigma''}\omega\ .\tag{1}$$
Now divide the common parameter interval $[0,1]$ into  sufficiently   small pieces $[t_{k-1},t_k]$ $\ (1\leq k\leq N)$, such that to each piece the equality $(1)$ can be applied. Summing the $N$ resulting equations up, the right sides telescope to $0$, so that we obtain
$$\int_{\gamma_0}\omega - \int_{\gamma_1}\omega=0\ ,$$
as stated.
The example with the two circles is a special case of the above.
[1] http://tinyurl.com/at9zypv
