Obtain the equation of the hypotenuse

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Context Panel: Student Precalculus≻
Lines And Segments≻Line≻Equation

$\left[0\,1\right]\,\left[1\,0\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-x\+1$ 

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Calculus - Multivariate≻Integration≻Center of Mass≻Cartesian 2-D

Table 6.5.8(a)   Solution by Task Template

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

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

Obtain $m$, the total mass of the lamina

${\int }_0^1{\int }_0^{1-x}\mathrm{\ρ} \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{1}{3}$$\stackrel{\text{assign to a name}}{\to }$$m$

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

${\int }_0^1{\int }_0^{1-x}\mathrm{\ρ}\cdot y \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{2}{15}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Mx}$

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

${\int }_0^1{\int }_0^{1-x}\mathrm{\ρ}\cdot x \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{2}{15}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{My}$

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

$\mathrm{My}\/m$ = $\frac{2}{5}$

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

$\mathrm{Mx}\/m$ = $\frac{2}{5}$

Table 6.5.8(b)   Calculation of the center of mass from first principles

Initialize

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

Obtain the equation of the hypotenuse of the right triangle

$\mathrm{Student}:-\mathrm{Precalculus}:-\mathrm{Line}\left(\left[0\,1\right]\,\left[1\,0\right]\right)\left[1\right]$

$y\=-x\+1$

Obtain the total mass of the lamina

$q≔\mathrm{Int}\left(\mathrm{\ρ}\,\left[y\=0..1-x\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^{-x\+1}\left(2x^2\+2y^2\right)\ⅆy\ⅆx$

$m≔\mathrm{value}\left(q\right)$

$\frac{1}{3}$

Obtain the first moments $M_x$ and $M_y$ 

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot y\,\left[y\=0..1-x\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^{-x\+1}\left(2x^2\+2y^2\right)y\ⅆy\ⅆx$

$M_x≔\mathrm{value}\left(q\right)$

$\frac{2}{15}$

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot x\,\left[y\=0..1-x\,x\=0..1\right]\right)$

${\∫}_0^1{\∫}_0^{-x\+1}\left(2x^2\+2y^2\right)x\ⅆy\ⅆx$

$M_y≔\mathrm{value}\left(q\right)$

$\frac{2}{15}$

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

$\left[\frac{M_y}{m}\,\frac{M_x}{m}\right]$ = $\left[\frac{2}{5}\,\frac{2}{5}\right]$

Obtain the inert integrals defining the center of mass

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\ρ}\,y\=0..1-x\,x\=0..1\,\mathrm{output}\=\mathrm{integral}\right)$

$\frac{{\∫}_0^1{\∫}_0^{1-x}\left(2x^2\+2y^2\right)x\ⅆy\ⅆx}{{\∫}_0^1{\∫}_0^{1-x}\left(2x^2\+2y^2\right)\ⅆy\ⅆx}\,\frac{{\∫}_0^1{\∫}_0^{1-x}\left(2x^2\+2y^2\right)y\ⅆy\ⅆx}{{\∫}_0^1{\∫}_0^{1-x}\left(2x^2\+2y^2\right)\ⅆy\ⅆx}$

Obtain the coordinates of the center of mass

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\ρ}\,y\=0..1-x\,x\=0..1\right)$

$\frac{2}{5}\,\frac{2}{5}$

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

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{CenterOfMass}\left(\mathrm{\ρ}\,y\=0..1-x\,x\=0..1\,\mathrm{output}\=\mathrm{plot}\,\mathrm{caption}\=''''\,\mathrm{scaling}\=\mathrm{constrained}\,\mathrm{labels}\=\left[x\,y\,z\right]\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{orientation}\=\left[-20\,70\,0\right]\,\mathrm{tickmarks}\=\left[2\,2\,3\right]\right)$

Table 6.5.8(c)   Coordinates of the center of mass calculated with the CenterOfMass command