I am trying to compute the line integral $\int_C \omega$, where $\omega=-y\sqrt{x^2+y^2}dx+x\sqrt{x^2+y^2}dy$ and $C$ is the circle $x^2+y^2=2x$ directly and using Green's Theorem.
I have managed to do it directly by parametrizing the circle: $x^2+y^2=2x,\ x=r\cos(\theta),\ y=r\sin(\theta)\Rightarrow x^2+y^2=r^2$ and $2x=2r\cos(\theta)$ so $x^2+y^2=2x$ becomes $r^2=2r\cos(\theta)$ hence $r=2\cos(\theta),\ x=r\cos(\theta)=2\cos^2(\theta)=2\cdot\left(\frac{1+\cos(2\theta)}{2}\right)=1+\cos(2\theta)\ y=r\sin(\theta)=2\cos(\theta)\sin(\theta)=\sin(2\theta);\ 0\leq \theta\leq \pi$ thus $\int_{C}\omega=\int_{\theta=0}^{\theta=\pi}(-\sin(2\theta)\sqrt{2+2\cos(2\theta)}\left(-2\sin(2\theta))+(1+\cos(2\theta))\sqrt{2+2\cos(2\theta)}2\cos(2\theta)\right)d\theta=\int_{0}^{\pi}\left(2\sin^2(2\theta)\sqrt{2\cdot(1+\cos(2\theta)}+(2\cos(2\theta)+2\cos^2(2\theta)\sqrt{2(1+\cos(2\theta)}\right)d\theta=\dots=\frac{32}{3}.$
Now, when I try to do it using Green's theorem I get (setting $D=\{(x,y)\in\mathbb{R}^2:(x-1)^2+y^2< 1\}$: $\int_{C}\omega=\int_{D}\text{d}\omega=\int_{D}3\sqrt{x^2+y^2}dxdy$ which is equal, using polar coordinates $x=1+r\cos(\theta),\ y=r\sin(\theta)$ to $3\int_{\theta=0}^{\theta=2\pi}\left(\int_{r=0}^{r=1}\sqrt{r^2+2r\cos(\theta)+1}rdr\right)d\theta$, an integral which, up to now, I haven't been able to evaluate so is there another (easier) way to set up this double integral or, alternatively, to evaluate the double integral I have already set up? Thanks.
The circle $C$ given by $(x-1)^2+y^2=1$ has polar equation $r(r-2\cos(\theta))=0$ or simply $r=2\cos(\theta).$ Using the usual polar coordinates, your integral becomes $$\int_{-\frac{\pi}2}^{\frac {\pi}2}\int_0^{2\cos(\theta)}3r^2~drd\theta.$$ The rest should be straightforward.