In Exercises, let $\mathcal{S}=\left\{(x,y)\mid x^2+y^2\le 1\text{and}x\ge 0\right\}.$

If the density at each point in S is proportional to the point’s distance from the origin, find the center of mass of S.

The given region is $\mathcal{S}=\left\{(x,y)\mid x^2+y^2\le 1\text{and}x\ge 0\right\}.$

the density at each point in S is proportional to the point’s distance from the origin

the density at each point in the region is proportional to the point’s distance from the origin, so the density function of the region is following.

$\rho (x,y)=k\sqrt{(x-0{)}^2+(y-0{)}^2}$

Substitute density function in the formula of the center of mass $\overline{x}.$

role="math" localid="1650317184961" $$\begin{gathered} \overline{x}=\frac{{\iint }_{\Omega }x\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA} \\ \overline{x}=\frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xk\sqrt{x^2+y^2}dydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx} \\ \overline{x}=\frac{2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kx\sqrt{x^2+y^2}dydx}{2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx} \\ \overline{x}=\frac{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kx\sqrt{x^2+y^2}dydx}{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx} \end{gathered}$$

Similarly center of mass $\overline{y}$is following.

$\overline{y}=\frac{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}ky\sqrt{x^2+y^2}dydx}{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx}$

Change the Cartesian coordinates system into the Polar coordinates system using relations $x=r\cos \theta ,y=r\sin \theta .$

$$\begin{gathered} \overline{x}=\frac{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kx\sqrt{x^2+y^2}dydx}{{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx} \\ \begin{array}{rl}\overline{x}& =\frac{{\int }_0^{\pi /2}{\int }_0^{1}k\left(r\cos \theta \right)\left(r\right)rdrd\theta }{{\int }_0^{\pi /2}{\int }_0^1\left(k\right)\left(r\right)rdrd\theta }\\ \overline{x}& =\frac{{\int }_0^{\pi /2}{\int }_0^1{r}^3\cos \theta drd\theta }{{\int }_0^{\pi /2}{\int }_0^1{r}^{2}drd\theta }\\ \overline{x}& =\frac{{\int }_0^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^1\cos \theta d\theta }{{\int }_0^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^{1}d\theta }\\ \overline{x}& =\frac{\frac{1}{4}{\int }_0^{\pi /2}\cos \theta d\theta }{\frac{1}{3}{\int }_0^{\pi /2}d\theta }\\ \overline{x}=& \frac{\frac{1}{4}{\int }_0^{\pi /2}\cos \theta d\theta }{\frac{1}{3}{\int }_0^{\pi /2}d\theta }\\ \overline{x}=& \frac{\frac{1}{4}[\sin \theta {]}_0^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}{\frac{1}{3}[\theta {]}_0^{\frac{\mathrm{\pi }}{2}}}\\ \overline{x}=& \frac{\frac{1}{4}\left(1\right)}{\frac{1}{3}\left(\frac{\mathrm{\pi }}{2}\right)}\\ \overline{x}=& \frac{3}{2\mathrm{\pi }}\end{array} \end{gathered}$$

The Center of mass$\overline{y}$is following.

role="math" localid="1650317672656" $$\begin{gathered} \overline{y}=\frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ky\sqrt{x^2+y^2}dydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx} \\ \begin{array}{rl}\overline{y}& =\frac{{\int }_{-\pi /2}^{\pi /2}{\int }_0^{1}k\left(r\sin \theta \right)\left(r\right)rdrd\theta }{{\int }_{-\pi /2}^{\pi /2}{\int }_0^1\left(k\right)\left(r\right)rdrd\theta }\\ y& =\frac{{\int }_{-\pi /2}^{\pi /2}{\int }_0^1{r}^3\sin \theta drd\theta }{{\int }_{-\pi /2}^{\pi /2}{\int }_0^1{r}^{2}drd\theta }\\ \overline{y}& =\frac{{\int }_{-\pi /2}^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^1\sin \theta d\theta }{{\int }_{-\pi /2}^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^{1}d\theta }\\ \overline{y}& =\frac{\frac{1}{4}{\int }_{-\pi /2}^{\pi /2}\sin \theta d\theta }{\frac{1}{3}{\int }_{-\pi /2}^{\pi /2}d\theta }\\ \overline{y}=& \frac{\frac{1}{4}{\int }_{-\pi /2}^{\pi /2}\sin \theta d\theta }{\frac{1}{3}{\int }_{-\pi /2}^{\pi /2}d\theta }\\ \overline{y}=& \frac{\frac{1}{4}[-\cos \theta {]}_{-\pi /2}^{{\displaystyle \frac{\mathrm{\pi }}{2}}}}{\frac{1}{3}[\theta {]}_{-\pi /2}^{\frac{\mathrm{\pi }}{2}}}\\ \overline{y}=& 0\end{array} \end{gathered}$$

So the center of mass of Sis$\left(\overline{x},\overline{y}\right)=\left(\frac{3}{2\mathrm{\pi }},0\right).$