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

$I_x\={\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot y^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{63}{2}$ 

$I_y\={\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot x^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{488}{5}$

$R_x\=\sqrt{\frac{I_y}{m}}\=\sqrt{\frac{488\/5}{25}}\=\frac{2}{25}\sqrt{610}$

$R_y\=\sqrt{\frac{I_x}{m}}$ =$\sqrt{\frac{63\/2}{25}}\=\frac{3}{10}\sqrt{14}$ 

Table 6.6.4(a)   Moments of inertia and radii of gyration

Obtain the equation of the hypotenuse

Loading Student:-Precalculus

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

Define the density $\mathrm{\ρ}$ 

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

Obtain the total mass $m$ 

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

Obtain the second moments

${\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot y^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{63}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Ix}$$$$$

${\int }_0^4{\int }_0^{3\left(1-x\/4\right)}\mathrm{\ρ}\cdot x^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{488}{5}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Iy}$$$$$

Obtain the radii of gyration

$R_x\=\sqrt{\mathrm{Iy}\/m}$

$R_x\=\frac{2}{25}\sqrt{610}$

$\stackrel{\text{at 10 digits}}{\to }$

$R_x\=1.975854246$

$R_y\=\sqrt{\mathrm{Ix}\/m}$

$R_y\=\frac{3}{10}\sqrt{14}$

$\stackrel{\text{at 10 digits}}{\to }$

$R_y\=1.122497216$

Initialize

$\mathrm{\ρ}≔1\+2 x\+y\/2\:$

Obtain the equation of the hypotenuse

Use the Line command from the Student Precalculus package.

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

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

Obtain the total mass $m$ 

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

$25$

Obtain the second moments

$\mathrm{Ix}≔\mathrm{int}\left(\mathrm{\ρ}\cdot y^2\,\left[y\=0..3\left(1-x\/4\right)\,x\=0..4\right]\right)$

$\frac{63}{2}$

$\mathrm{Iy}≔\mathrm{int}\left(\mathrm{\ρ}\cdot x^2\,\left[y\=0..3\left(1-x\/4\right)\,x\=0..4\right]\right)$

$\frac{488}{5}$

Obtain the radii of gyration

$$\begin{gathered} R_x\=\mathrm{sqrt}\left(\frac{\mathrm{Iy}}{m}\right)\; \\ {R}_x\=\mathrm{evalf}\left(\mathrm{sqrt}\left(\frac{\mathrm{Iy}}{m}\right)\right) \end{gathered}$$

$R_x\=\frac{2}{25}\sqrt{610}$

$$\begin{gathered} R_y\=\mathrm{sqrt}\left(\frac{\mathrm{Ix}}{m}\right)\; \\ {R}_y\=\mathrm{evalf}\left(\mathrm{sqrt}\left(\frac{\mathrm{Ix}}{m}\right)\right) \end{gathered}$$

$R_y\=\frac{3}{10}\sqrt{14}$