Question

Let $S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

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

Short answer

Answer is$\left(\frac{3\pi }{16},\frac{3}{8}\right)$

Step-by-step solution

Step 1. Given information

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

Step 2. Explanation

$S=\left\{\left(x,y\right)|x^2+y^2\le 1,x\ge 0\right\}$

$\begin{array}{rcl}\overline{x}& =& \frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xkxdydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}kxdydx}\\ & =& \frac{2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}k{x}^{2}dydx}{2{\int }_0^1{\int }_0^{\sqrt{1-x^2}}kxdydx}\end{array}$

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

width="336">x¯=∫0π/2∫01kr2cos2θrdrdθ∫0π/2∫01krcosθrdrdθ=∫0π/2∫01r3cos2θdrdθ∫0π/2∫01r2cosθdrdθ=∫0π/2r4401cos2θdθ∫0π/2r3301cosθdθ=∫0π/2r4401cos2θdθ∫0π/2r3301cosθdθ=14∫0π/2cos2θdθ13∫0π/2cosθdθusecos2θ=1+cos2θ2=14∫0π/212(1+cos2θ)dθ13∫0π/2cosθdθ=18θ+12sin2θ0π/213[sinθ]0π/2=π16-sin2π213sinπ2-sin(0)=π16-013(1-0)=3π16

$\begin{array}{rcl}\overline{y}& =& \frac{{\iint }_{\Omega }y\rho (x,y)dA}{{\iint }_{\Omega }\rho (x,y)dA}\\ & =& \frac{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}ykxdydx}{{\int }_0^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}kxdydx}\\ & =& \frac{{\int }_0^{\pi /2}{\int }_0^{1}r\sin \theta kr\cos \theta rdrd\theta }{{\int }_0^{\pi /2}{\int }_0^{1}kr\cos \theta rdrd\theta }\\ & =& \frac{{\int }_0^{\pi /2}{\int }_0^1{r}^3\sin \theta \cos \theta drd\theta }{{\int }_0^{\pi /2}{\int }_0^1{r}^2\cos \theta drd\theta }\\ & =& \frac{\frac{1}{2}{\int }_0^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^1\sin 2\theta d\theta }{{\int }_0^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^1\cos \theta d\theta }\\ & =& \frac{\frac{1}{8}{\int }_0^{\pi /2}\sin 2\theta d\theta }{\frac{1}{3}{\int }_0^{\pi /2}\cos \theta d\theta }\\ & =& \frac{\frac{1}{8}{\left[-\frac{1}{2}\cos 2\theta \right]}_0^{\pi /2}}{\frac{1}{3}[\sin \theta {]}_0^{\pi /2}}\\ & =& \frac{\frac{1}{8}\left[-\frac{1}{2}\cos 2\left(\frac{\pi }{2}\right)+\frac{1}{2}\cos \left(0\right)\right.}{\frac{1}{3}\left[\sin \left(\frac{\pi }{2}\right)-\sin \left(0\right)\right]}\\ & =& \frac{\left(\frac{1}{8}\right)}{\left(\frac{1}{3}\right)}\\ & =& \frac{3}{8}\end{array}$

Therefore, the center of mass of the region is $(\overline{x},\overline{y})=\left(\frac{3\pi }{16},\frac{3}{8}\right)$