Finding a length of arc, what's wrong? 
Find: $$ \int \sqrt{x^{2}+y^{2}}dl$$
  $$L: x^{2}+y^{2}= Rx$$


(at image $p' = -R\cdot \sin(\phi)$ )
 A: Hints:
$$x^2+y^2=Rx\implies \left(x-\frac R2\right)^2+y^2=\frac{R^2}4$$
If you want/can you can make a translation to get a canonical circle of radius $\,\frac R2\;$ , so its parametrization (which is where you went wrong) would be
$$x(t)=\frac R2\cos t\;,\;\;y(t)=\frac R2\sin t\;,\;\;0\le t\le 2\pi\implies$$
$$l=\int\limits_0^{2\pi}\sqrt{x'(t)^2+y'(t)^2}dt=\int\limits_0^{2\pi}\sqrt{\frac{R^2}4\sin^2t+\frac{R^2}4\cos^2t}dt=$$
$$=\frac R2\int\limits_0^{2\pi}dt=\pi R$$
A: As a caution, I thought I should mention something about this curve by using this variant approach.  You can translate the equation as it stands into polar coordinates as
$$ x^2 \ + \ y^2 \ = \ Rx \ \rightarrow \ r^2 \ = \ R \cdot r \cos \theta\ \Rightarrow \ r \ = \ R \cos \theta \ . $$
The arclength integral in polar coordinates would be
$$\int_{\theta_1}^{\theta_2}  \sqrt{ \ r^2 \ + \ (\frac{dr}{d\theta})^2 } \ \ d\theta \ = \ \int_{\theta_1}^{\theta_2}  \sqrt{ \ R^2 \cos^2 \theta \ + \ R^2 \sin^2 \theta } \ \ d\theta \ = \ \int_{\theta_1}^{\theta_2}  R\ \ d\theta \ . $$
The caution is this: while the circle which DonAntonio uses, which has been translated to have its center at the origin, can be integrated fully from $ \ 0 \ $ to $ \ 2 \pi \ $ in the angle variable, the "circle" described here is in fact a "one-petal" rosette.  While it has the shape of a circle, it is actually traced once completely by taking the angle variable from $ \ 0 \ $ to $ \ \pi \ . $  Thus, the arclength integration here needs to be
$$  \int_{0}^{\pi}  R\ \ d\theta \ = \ \pi R \ , $$
which is sensible, since  $ \ R \ $ in the equation for this polar curve is actually the diameter of the "circle".
A: You want to evaluate
$$\oint_L d\ell \sqrt{x^2+y^2}$$
where
$$\left ( x-\frac{R}{2}\right)^2+y^2=\frac{R^2}{4}$$
Parametrizing:
$$x(t) = \frac{R}{2} + \frac{R}{2} \cos{t} = R\cos^2{(t/2)}$$
$$y(t)=\frac{R}{2} \sin{t}$$
$$d\ell = \sqrt{x'^2+y'^2} dt = \frac{R}{2} dt$$
$$\sqrt{x^2+y^2} = \sqrt{R x} = R |\cos{(t/2)}|$$
Then the integral is equal to
$$\frac{R}{2} R \int_0^{2 \pi} dt\, |\cos{(t/2)}| = R^2 \int_0^{\pi} dt \, \cos{(t/2)} = 2 R^2$$
