Question

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.

Short answer

The Center of mass of regionSis$\left(\overline{x},\overline{y}\right)=\left(\frac{3}{2\mathrm{\pi }},0\right).$

Step-by-step solution

Step 1. Given information.

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

Step 2. The formula for the center of mass 

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.

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

Step 3. x-coordinate of Center of mass x.

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

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

Step 4. y-coordinate of Center of mass y

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$\left(\overline{x},\overline{y}\right)=\left(\frac{3}{2\mathrm{\pi }},0\right).$