Q 47. - Calculus

Question

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

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

Step-by-Step Solution

Answer

Answer is $\left(0,0\right)$

Step 1. Given information

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

Step 2. Explanation

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

$\begin{array}{rcl}\overline{x}& =& \frac{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} x kx dy dx}{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} kx dy dx}\\ & =& \frac{4{\int }_0^1{\int }_0^{\sqrt{1-x^2}} x kx dy dx}{4{\int }_0^1{\int }_1^{\sqrt{1-x^2}} kx dy dx}\\ & =& \frac{{\int }_0^{1}k{x}^2{\left[y\right]}_0^{\sqrt{1-x^2}} dy dx}{{\int }_0^1 kx {\left[y\right]}_0^{\sqrt{1-x^2}}dy dx}\\ & =& \frac{{\int }_0^{1}k{x}^2\sqrt{1-x^2} dy dx}{{\int }_0^1 kx \sqrt{1-x^2}dy dx}\\ & =& \frac{{\displaystyle \frac{1}{8}}{\left[x\sqrt{1-x^2}\left(2x^2-1\right)+{\sin }^{-1}x\right]}_0^1}{{\displaystyle \frac{4}{3}}}\\ & =& \frac{{\displaystyle 0}}{{\displaystyle \frac{4}{3}}}\\ & =& 0\\ & & \end{array}$

$\begin{array}{rcl}\overline{y}& =& \frac{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y kx dy dx}{{\int }_{-1}^1{\int }_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} kx dy dx}\\ & =& \frac{{\int }_{-1}^1\mathrm{kx}{\left[{\displaystyle \frac{{\mathrm{y}}^2}{2}}\right]}_{-\sqrt{1-{\mathrm{x}}^2}}^{\sqrt{1-{\mathrm{x}}^2}}}{{\int }_{-1}^1\mathrm{kx}{\left[\mathrm{y}\right]}_{-\sqrt{1-{\mathrm{x}}^2}}^{\sqrt{1-{\mathrm{x}}^2}}}\\ & =& 0\\ & & \end{array}$