Question

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

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

Short answer

Answer is (0,0)

Step-by-step solution

Step 1. Given information

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

Step 2. Explanation

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

$\begin{array}{rcl}\overline{x}& =& \frac{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}xk\sqrt{x^2+y^2}dydx}{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx}\\ \overline{x}& =& \frac{2{\int }_{-1}^1{\left[\frac{1}{2}y\sqrt{x^2+y^2}+\frac{1}{2}{x}^2\ln \left(y+\sqrt{x^2+y^2}\right)\right]}_0^{\sqrt{1-x^2}}xdx}{2{\int }_{-1}^1{\left[\frac{1}{2}y\sqrt{x^2+y^2}+\frac{1}{2}{x}^2\ln \left(y+\sqrt{x^2+y^2}\right)\right]}_0^{\sqrt{1-x^2}}dx}\\ \overline{x}& =& \frac{2{\int }_{-1}^1\left[\frac{1}{2}\sqrt{1-x^2}+\frac{1}{2}{x}^2\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)\right]xdx}{2{\int }_{-1}^1\left[\frac{1}{2}\sqrt{1-x^2}+\frac{1}{2}{x}^2\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)\right]dx}\\ & =& 0\\ \overline{y}& =& \frac{{\displaystyle {\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}yk\sqrt{x^2+y^2}dydx}}{{\displaystyle {\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}k\sqrt{x^2+y^2}dydx}}\\ & =& \frac{{\displaystyle 0}}{{\displaystyle \frac{{\displaystyle 3}}{{\displaystyle 2}}\pi }}\\ & =& 0\end{array}$