Center of Mass - Unbalanced Ring

This is a different approach.

We could write $\displaystyle \mathbf{r}_{CM} =\displaystyle \frac{\displaystyle \int \mathbf{r} d \mathbf{m}}{ \displaystyle \int d\mathbf{m}}$ The points on the ring can be written as $\mathbf{r} = e^{i \theta}$ where $0 \le \theta \le 2 \pi$ . Hence

$\displaystyle \mathbf{r}_{CM} = \frac{\displaystyle \int_0^{2 \pi} e^{i \theta} e^{- \theta} d \theta}{\displaystyle\int_0^{2 \pi} e^{- \theta} d \theta}$

$\implies \displaystyle \mathbf{r}_{CM} = \frac{\displaystyle \int_0^{2 \pi} e^{ \theta ( i -1)}}{\displaystyle -e^{- \theta} {\huge|}_{0}^{2 \pi}}$

$\implies \mathbf{r}_{CM} = \displaystyle \frac{ \displaystyle \frac{1}{i -1} \displaystyle e^{\theta (i-1)} {\huge|}_0^{2 \pi}}{ 1- e^{-2 \pi } }$

$\implies \mathbf{r}_{CM} = \displaystyle \frac{ \displaystyle \frac{1}{i -1} \left( e^{2 \pi (i -1)} - 1 \right) }{ 1- e^{-2 \pi } }$

$\implies \mathbf{r}_{CM} = \displaystyle \frac{ \displaystyle \frac{1}{i -1} \left( e^{2 \pi i} e^{-2 \pi} - 1 \right) }{ 1- e^{-2 \pi } }$

$\implies \mathbf{r}_{CM} = \displaystyle \frac{ \displaystyle \frac{1}{i -1} \left( e^{-2 \pi} - 1 \right) }{ 1- e^{-2 \pi } }$

$\implies \mathbf{r}_{CM} = \frac{(-1)}{i-1 }$

$\implies \mathbf{r}_{CM} = \frac{i + 1}{2}$

$\| \mathbf{r}_{CM} \| = \frac{1}{\sqrt{2}}$

P.S I am a little new to $\LaTeX$

The $x$ and $y$ coordinates of the center of mass of the ring are respectively

$x_{CM} = \dfrac{\displaystyle \int x \,dm}{\displaystyle \int \,dm} \qquad , \qquad y_{CM} = \dfrac{\displaystyle \int y \,dm}{\displaystyle \int \,dm}$

Plugging in all terms in terms of the parameter $\theta$ and the limits of integral from $0$ to $2\pi$ , we have

$\begin{aligned} & x_{CM} = \dfrac{\displaystyle \int_0^{2\pi} \cos \theta \dfrac{dm}{d\theta} \,d\theta}{\displaystyle \int_0^{2\pi} \dfrac{dm}{d\theta} \,d\theta} &,& y_{CM} = \dfrac{\displaystyle \int_0^{2\pi} \sin \theta \dfrac{dm}{d\theta} \,d\theta}{\displaystyle \int_0^{2\pi} \dfrac{dm}{d\theta} \,d\theta} \\ \implies & x_{CM} = \dfrac{\displaystyle \int_0^{2\pi} \cos \theta e^{-\theta} \,d\theta}{\displaystyle \int_0^{2\pi} e^{-\theta} \,d\theta} &,& y_{CM} = \dfrac{\displaystyle \int_0^{2\pi} \sin \theta e^{-\theta} \,d\theta}{\displaystyle \int_0^{2\pi} e^{-\theta} \,d\theta} \\ \implies & x_{CM} = \dfrac{1}{2} \cdot \dfrac{-e^{-\theta} (\cos\theta + \sin\theta) {\huge|}_0^{2\pi}}{-e^{-\theta} {\huge|}_0^{2\pi}} &,& y_{CM} = \dfrac{1}{2} \cdot \dfrac{e^{-\theta} (\sin\theta - \cos\theta) {\huge|}_0^{2\pi}}{-e^{-\theta} {\huge|}_0^{2\pi}} \\ \implies & x_{CM} = \dfrac12 &,& y_{CM} = \dfrac12 \end{aligned}$

Required distance of center of mass from the origin is thus, $r = \sqrt{(x_{CM})^2 + (y_{CM})^2} = \dfrac{1}{\sqrt{2}} = \boxed{0.707}$ correct to 3 decimal places.