Find the mass $M$ and the center of mass $(\bar{x},\bar{y})$ of a lamina in the shape of a region $R$ in the $xy$-plane that lies outside the circle $x^{2}+y^{2}=a^{2}$ and inside the circle $x^{2}+y^{2}=2ax.$ The mass density $\rho$ is inversely proportional to the distance from the origin.

Figure 34 $a\leq r\leq 2a\cos\theta$, $-\dfrac{\pi}{3}\leq \theta\leq \dfrac{\pi}{3}$

Solution Figure 34 illustrates the region $R$. Since $R$ involves two circles, we use polar coordinates. Then the two circles are given by $r=a$ and by $r=2a\cos \theta$. The circles intersect when $a=2a\cos \theta$ or $\cos \theta =\dfrac{1}{2},$ and the points of intersection are $\left( a,\dfrac{\pi }{3}\right)$ and $\left( a,-\dfrac{\pi }{3}\right) .$ The region $R$ is given by $a\leq r\leq 2a\cos \theta ,$ $-\dfrac{\pi }{3}\leq \theta \leq \dfrac{\pi }{3}.$

Since the distance of a point $(x,y)$ from the origin is $\sqrt{x^{2}+y^{2}}$ , the mass density of the lamina at any point $(x,y)$ in $R$ is $$\rho (x,y)=\dfrac{k}{\sqrt{x^{2}+y^{2}}}=\dfrac{k}{r} \qquad {\color{#0066A7}{\hbox{\(r^{2} =x^{2} + y^{2}\)}}}$$

where $k$ is the constant of proportionality. Since $R$ does not contain the origin, the mass density $\rho =\rho (x,y)$ is continuous on $R.$ The mass $M$ of the lamina is $$\begin{eqnarray*} M &=&\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}=k\displaystyle\iint\limits_{\kern-3ptR}\dfrac{rdr\,d\theta }{r} =k\int_{-\pi /3}^{\pi /3}\int_{a}^{2a\cos \theta }\,dr\,d\theta \\[4pt] &=&k \int_{-\pi /3}^{\pi /3}(2a\cos \theta -a)\,d\theta =k\big[2a\sin \theta -a\theta \big] _{-\pi /3}^{\pi /3}= k\!\left( 2a\sqrt{3}-a\dfrac{2\pi }{3}\right)\\[4pt] &=&\dfrac{2ka}{3}(3\sqrt{3}-\pi ) \end{eqnarray*}$$

Both the mass density function and the region $R$ are symmetric about the $x$-axis. So, the center of mass lies on the $x$-axis; that is, $\bar{y}=0$. To find $\bar{x}$, we first find $M_{y}$. $$\begin{eqnarray*} M_{y}& =&\displaystyle\iint\limits_{\kern-3ptR}x\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}( r\cos \theta ) \left( \dfrac{k}{r}\right) r\,dr\,d\theta =k\int_{-\pi /3}^{\pi /3}\int_{a}^{2a\cos \theta }r\cos \theta \,dr\,d\theta\\[4pt] &=&\dfrac{k}{2}\int_{-\pi /3}^{\pi /3}(4a^{2}\cos ^{2}\theta -a^{2})\cos \theta \,d\theta = a^{2}\dfrac{k}{2}\int_{-\pi /3}^{\pi /3}(3-4\sin ^{2}\theta )\cos \theta \,d\theta\\[4pt] &=&\dfrac{a^{2}k}{2}\left[ 3\sin \theta -\dfrac{4\sin ^{3}\theta }{3} \right] _{-\pi /3}^{\pi /3}=ka^{2}\sqrt{3} \end{eqnarray*}$$

Then $$\bar{x}=\dfrac{M_{y}}{M}=\dfrac{ka^{2}\sqrt{3}}{\dfrac{2ka}{3}(3\sqrt{3}-\pi )} =\dfrac{3a\sqrt{3}}{2(3\sqrt{3}-\pi )}\approx 1.265a$$

The center of mass is approximately $(1.265a,0)$.