Q 54. - Calculus

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.

Step-by-Step Solution

Answer

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

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" style="max-width: none; vertical-align: -442px;" $\begin{array}{rcl}\overline{x}& =& \frac{{\int }_0^{\pi /2}{\int }_0^{1}k{r}^2{\cos }^2\theta rdrd\theta }{{\int }_0^{\pi /2}{\int }_0^{1}kr\cos \theta rdrd\theta }\\ & =& \frac{{\int }_0^{\pi /2}{\int }_0^1{r}^3{\cos }^2\theta drd\theta }{{\int }_0^{\pi /2}{\int }_0^1{r}^2\cos \theta drd\theta }\\ & =& \frac{{\int }_0^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^1{\cos }^2\theta d\theta }{{\int }_0^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^1\cos \theta d\theta }\\ & =& \frac{{\int }_0^{\pi /2}{\left[\frac{r^4}{4}\right]}_0^1{\cos }^2\theta d\theta }{{\int }_0^{\pi /2}{\left[\frac{r^3}{3}\right]}_0^1\cos \theta d\theta }\\ & =& \frac{\frac{1}{4}{\int }_0^{\pi /2}{\cos }^2\theta d\theta }{\frac{1}{3}{\int }_0^{\pi /2}\cos \theta d\theta }\left(\text{ use }{\cos }^2\theta =\frac{1+\cos 2\theta }{2}\right)\\ & =& \frac{\frac{1}{4}{\int }_0^{\pi /2}\frac{1}{2}(1+\cos 2\theta )d\theta }{\frac{1}{3}{\int }_0^{\pi /2}\cos \theta d\theta }\\ & =& \frac{\frac{1}{8}{\left[\theta +\frac{1}{2}\sin 2\theta \right]}_0^{\pi /2}}{\frac{1}{3}[\sin \theta {]}_0^{\pi /2}}\\ & =& \frac{\frac{\pi }{16}-\sin 2\left(\frac{\pi }{2}\right)}{\frac{1}{3}\left(\sin \left(\frac{\pi }{2}\right)-\sin \left(0\right)\right)}\\ & =& \frac{\frac{\pi }{16}-0}{\frac{1}{3}(1-0)}\\ & =& \frac{3\pi }{16}\end{array}$

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

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