Obtain the equation of the hypotenuse

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

$\left[0\,3\right]\,\left[4\,0\right]$$\stackrel{\text{equation of line}}{\to }$$y\=-\frac{3}{4}x\+3$

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

Table 6.5.4(a)   Solution by Task Template

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

$\mathrm{\ρ}\=1\+2 x\+ y\/2$$\stackrel{\text{assign}}{\to }$

Obtain $m$, the total mass of the lamina

${\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ} \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $25$$\stackrel{\text{assign to a name}}{\to }$$m$

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

${\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot y \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{45}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Mx}$

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

${\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot x \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $43$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{My}$

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

$\mathrm{My}\/m$ = $\frac{43}{25}$

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

$\mathrm{Mx}\/m$ = $\frac{9}{10}$

Table 6.5.4(b)   Calculation of the center of mass from first principles

Initialize

$\mathrm{\ρ}≔1\+2 x\+ y\/2\:$

Obtain the equation of the hypotenuse of the right triangle

$\mathrm{Student}:-\mathrm{Precalculus}:-\mathrm{Line}\left(\left[0\,3\right]\,\left[4\,0\right]\right)\left[1\right]$

$y\=-\frac{3}{4}x\+3$

Obtain the total mass of the lamina

$q≔\mathrm{Int}\left(\mathrm{\ρ}\,\left[y\=0..3\left(1-x\/4\right)\,x\=0..4\right]\right)$

${\∫}_0^4{\∫}_0^{-\frac{3}{4}x\+3}\left(1\+2x\+\frac{1}{2}y\right)\ⅆy\ⅆx$

$m≔\mathrm{value}\left(q\right)$

$25$

Obtain the first moments $M_x$ and $M_y$ 

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot y\,\left[y\=0..3\left(1-x\/4\right)\,x\=0..4\right]\right)$

${\∫}_0^4{\∫}_0^{-\frac{3}{4}x\+3}y\left(1\+2x\+\frac{1}{2}y\right)\ⅆy\ⅆx$

$M_x≔\mathrm{value}\left(q\right)$

$\frac{45}{2}$

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot x\,\left[y\=0..3\left(1-x\/4\right)\,x\=0..4\right]\right)$

${\∫}_0^4{\∫}_0^{-\frac{3}{4}x\+3}x\left(1\+2x\+\frac{1}{2}y\right)\ⅆy\ⅆx$

$M_y≔\mathrm{value}\left(q\right)$

$43$

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

$\left[\frac{M_y}{m}\,\frac{M_x}{m}\right]$ = $\left[\frac{43}{25}\,\frac{9}{10}\right]$