I am referring to this explanation of the polar planimeter, pp. 7
I think I do understand the following result:
During this operation, the wheel on the tracer arm will cover a distance of $$ \frac{b}{L}(x\cos\phi+y\sin\phi-b)d\phi+ad\theta. $$
But I am not sure I did understand why some of these terms seem to vanish when integrating over the curve $C$ and what is meant with $C_1'$ and $C_2'$:
My understanding would be that
$$ \oint_{C}d\phi + d\theta=0 $$ since this is a integral of a total differential $dF$ for $F(\phi,\theta)=\phi + \theta$ over a closed curve and the theory tells me that such integrals vanish. Hence it remains $$ \frac{b}{L}\oint_C(x\cos\phi+y\sin\phi)d\phi=\frac{b}{L}\oint_{C}x\cos\phi d\phi+\frac{b}{L}\oint_C y\sin\phi d\phi. $$
Now in the linked article we have instead
$$ \frac{b}{L}\oint_{C_1'}x\cos\phi d\phi+\frac{b}{L}\oint_{C_2'} y\sin\phi d\phi $$
and I am not sure what $C_1'$ and $C_2'$ are supposed to mean although it is said that $C_1'$ is the curve described by $(x,\phi)$ and $C_2'$ is the curve described by $(\phi,y)$
Maybe the only point that the author wanted to make is that the first integrand somehow depends on $x$ and $\phi$ whereas the second depends on $y$ and $\phi$; maybe thats all.
More formally, if $$ \gamma=(\gamma_1,\gamma_2)\colon [a,b]\to C $$ is the parametrization of curve $C$, then maybe the author wanted to say that we have something like $C=C_1'\cup C_2'$, i.e. a part of the curve which is expressed in terms of $x$ and $\phi$ and another which can be expressed in terms of $\phi$ and $y$.
That would be my explanation..