Centre of mass of sector. 
How do I find the centre of mass (COM) of a sector of angle $\alpha$?

Attempt: 
Go to an angle $\theta$ from $0$ rad and then choose a triangular strip of angle $d\theta$ and base $Rd\theta$. The centre of mass of this triangular strip lies at a height $\dfrac R3$ from the base. What do I do after this? How do I express the $R/3$ in terms of $\theta$? After that I just need to integrate from $0$ to $\alpha$  
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Hereafter, $\ds{d}$ is the sector radius:
\begin{align}
\vec{\mc{R}}_{cm} & \equiv
{\ds{\int_{sector}\vec{r}\,\dd^{2}\vec{r}} \over \ds{\int_{sector}\dd^{2}\vec{r}}} =
{\ds{\int_{0}^{\alpha}\int_{0}^{d}\bracks{r\cos\pars{\theta}\,\hat{x} +
r\sin\pars{\theta}\,\hat{y}}r\,\dd r\,\dd\theta} \over \ds{\int_{0}^{\alpha}\int_{0}^{d}r\,\dd r\,\dd\theta}}
\\[5mm] & =
{\ds{\pars{d^{3}/3}\braces{\bracks{%
\int_{0}^{\alpha}\cos\pars{\theta}\,\dd\theta}\hat{x} +
\bracks{\int_{0}^{\alpha}\sin\pars{\theta}\,\dd\theta}\hat{y}}} \over \ds{\pars{d^{2}/2}\int_{0}^{\alpha}\dd\theta}}
\\[5mm] & =
{2 \over 3}\,d\,{\sin\pars{\alpha}\,\hat{x} + 2\sin^{2}\pars{\alpha/2}\,\hat{y} \over \ds{\alpha}} =
\bbx{{2 \over 3}\,{\sin\pars{\alpha} \over \alpha}\,d\,\hat{x} +
{2 \over 3}\,{\sin\pars{\alpha/2} \over \alpha/2}\,d\,\hat{y}}
\end{align}
A: If you set up the sector $-\alpha/2 \le \theta \le \alpha/2$ and $0 \leq r \leq R$ then it is clear that the COM lies on the $x$-axis.  You need to find the $x$-coordinate.
$$
x_{\text{COM}} = \frac{\int_{\text{sector}} x\,dM}{\int_{\text{sector}} \,dM}
     = \frac{\int_{\text{sector}} x\,dA}{\int_{\text{sector}} \,dA}
     = \frac{\int_{r=0}^{R} \int_{\theta = -\alpha/2}^{\alpha/2} r^2\cos\theta\,d\theta dr}{R^2\alpha/2}
     =\frac{4R\sin(\alpha/2)}{3\alpha}
$$
