$m\={\int }_0^1{\int }_0^{1-x}\mathrm{\rho } \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $1\/3$

$I_x\={\int }_0^1{\int }_0^{1-x}{y}^2\cdot \mathrm{\ρ} \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $7\/90$

$I_y\={\int }_0^1{\int }_0^{1-x}{x}^2\cdot \mathrm{\ρ} \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $7\/90$

$R_x\=\sqrt{I_y\/m}\=\sqrt{\frac{7\/90}{1\/3}}$ = $\frac{\sqrt{210}}{30}$ ≐ 0.48

$R_y\=\sqrt{I_x\/m}\=$$\sqrt{\frac{7\/90}{1\/3}}$ = $\frac{\sqrt{210}}{30}$ ≐ 0.48

Table 6.6.8(a)   Moments of inertia and radii of gyration

Obtain the equation of the hypotenuse

Loading Student:-Precalculus

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

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 $I_x$, the moment of inertia about the $x$-axis

${\int }_0^1{\int }_0^{1-x}\mathrm{\ρ}\cdot y^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{7}{90}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Ix}$

Obtain $I_y$, the moment of inertia about the $y$-axis

${\int }_0^1{\int }_0^{1-x}\mathrm{\ρ}\cdot x^2 \mathit{\ⅆ}y \mathit{\ⅆ}x$ = $\frac{7}{90}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Iy}$$$

Obtain $R_x$

$\sqrt{\mathrm{Iy}\/m}$ = $\frac{1}{30}\sqrt{210}$$\stackrel{\text{at 5 digits}}{\to }$$0.48303$

Obtain $R_y$ 

$\sqrt{\mathrm{Ix}\/m}$ = $\frac{1}{30}\sqrt{210}$$\stackrel{\text{at 5 digits}}{\to }$$0.48303$

Table 6.5.8(b)   Moments of inertia and radii of gyration 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 moments of inertia $I_x$ and $I_y$ 

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot y^2\,\left[y\=0..1-x\,x\=0..1\right]\right)$

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

$\mathrm{Ix}≔\mathrm{value}\left(q\right)$

$\frac{7}{90}$

$q≔\mathrm{Int}\left(\mathrm{\ρ}\cdot x^2\,\left[y\=0..1-x\,x\=0..1\right]\right)$

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

$\mathrm{Iy}≔\mathrm{value}\left(q\right)$

$\frac{7}{90}$

Obtain the radii of gyration

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

$R_x\=\frac{1}{30}\sqrt{210}$

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

$R_y\=\frac{1}{30}\sqrt{210}$