Obtain the intersections of the bounding curves

$x\=y^2\,y\=x-2$

$x\=y^2\,y\=x-2$

$\stackrel{\text{solve}}{\to }$

$\left\{x\=1\,y\=-1\right\}\,\left\{x\=4\,y\=2\right\}$

Define the density function $\mathrm{\ρ}\left(x\,y\right)$ 

$\mathrm{\ρ}\=\left(1\+2 x^2\+3 y^2\right)\/10$$\stackrel{\text{assign}}{\to }$

Obtain $m$, the total mass of the lamina

${\int }_{-1}^2{\int }_{y^2}^{y\+2}\mathrm{\ρ} \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{6183}{1400}$$\stackrel{\text{assign to a name}}{\to }$$m$

Obtain $M_x$, the total moments about the $x$-axis

${\int }_{-1}^2{\int }_{y^2}^{y\+2}\mathrm{\ρ}\cdot y \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{108}{25}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Mx}$$$ 

Obtain $M_y$, the total moments about the $y$-axis

${\int }_{-1}^2{\int }_{y^2}^{y\+2}\mathrm{\ρ}\cdot x \mathit{\ⅆ}x \mathit{\ⅆ}y$ = $\frac{7263}{700}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{My}$$$

Obtain $\stackrel{\&conjugate0;}{x}\=M_y\/m$ 

$\mathrm{My}\/m$ = $\frac{538}{229}$$$

Obtain $\stackrel{\&conjugate0;}{y}\=M_x\/m$ 

$\mathrm{Mx}\/m$ = $\frac{224}{229}$$$

Table 6.5.7(a)   Calculation of the center of mass from first principles

Initialize

$\mathrm{\ρ}≔\left(1\+2 x^2\+ 3 y^2\right)\/10\:$

Obtain the total mass of the lamina

$q≔\mathrm{Int}\left(\mathrm{\ρ}\,\left[x\=y^2..y\+2\,y\=-1..2\right]\right)$

${\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)\ⅆx\ⅆy$

$m≔\mathrm{value}\left(q\right)$

$\frac{6183}{1400}$

Obtain the first moments $M_x$ and $M_y$ 

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot y\,\left[x\=y^2..y\+2\,y\=-1..2\right]\right)$

${\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)y\ⅆx\ⅆy$

$M_x≔\mathrm{value}\left(q\right)$

$\frac{108}{25}$

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot x\,\left[x\=y^2..y\+2\,y\=-1..2\right]\right)$

${\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)x\ⅆx\ⅆy$

$M_y≔\mathrm{value}\left(q\right)$

$\frac{7263}{700}$

Obtain the coordinates of the center of mass $\left(\stackrel{\&conjugate0;}{x}\,\stackrel{\&conjugate0;}{y}\right)$ 

$\left[\frac{M_y}{m}\,\frac{M_x}{m}\right]$ = $\left[\frac{538}{229}\,\frac{224}{229}\right]$$$

Table 6.5.7(b)   Coordinates of the center of mass calculated from first principles

Obtain the inert integrals defining the center of mass

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\ρ}\,x\=y^2..y\+2\,y\=-1..2\,\mathrm{output}\=\mathrm{integral}\right)$

$\frac{{\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)x\ⅆx\ⅆy}{{\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)\ⅆx\ⅆy}\,\frac{{\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)y\ⅆx\ⅆy}{{\∫}_{-1}^2{\∫}_{y^2}^{y\+2}\left(\frac{1}{5}{x}^2\+\frac{3}{10}{y}^2\+\frac{1}{10}\right)\ⅆx\ⅆy}$

Obtain the coordinates of the center of mass

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\rho }\,x\=y^2..y\+2\,y\=-1..2\right)$

$\frac{538}{229}\,\frac{224}{229}$

Obtain a graph of the region $R$, the density $\mathrm{\ρ}$, and the center of mass

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\rho }\,x\=y^2..y\+2\,y\=-1..2\,\mathrm{output}\=\mathrm{plot}\,\mathrm{caption}\=''''\,\mathrm{scaling}\=\mathrm{constrained}\,\mathrm{labels}\=\left[x\,y\,z\right]\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{orientation}\=\left[-110\,60\,0\right]\,\mathrm{tickmarks}\=\left[5\,3\,5\right]\right)$

Table 6.5.7(c)   Coordinates of the center of mass calculated with the CenterOfMass command