Q. 57 - Calculus

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.

Step-by-Step Solution

Answer

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

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 $\overline{x}.$

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

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 $x=r \cos \theta ,y=r \sin \theta .$

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

Step 4. y -coordinate of Center of mass y

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

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

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