91Ó°ÊÓ

Let the vertices of rectangular region is $(0,0),(b,0),(0,h)\text{and}(b,h)$

$\mathrm{ÒÏ}(x,y)=k{x}^2$

The formula is $I_y={âˆ\neg }_{\mathrm{Î\copyright }}{x}^2\mathrm{ÒÏ}(x,y)dA$

Put limits $I_y={∫}_0^b{∫}_0^h{x}^2\mathrm{ÒÏ}(x,y)dydx$

$I_y={∫}_0^b{∫}_0^h{x}^{2}k{x}^{2}dydx\left[\mathrm{ÒÏ}(x,y)=k{x}^2\right]$

$I_y=k{∫}_0^b{∫}_0^h{x}^{4}dydx$

Solving inner integral first

$I_y=k{∫}_0^b[y{]}_0^h{x}^{4}dx=k{∫}_0^b[h]x^{4}dx=kh{∫}_0^b{x}^{4}dx$

Solving further

$I_y=kh{\left[\frac{x^5}{5}\right]}_0^b$

$I_y=kh\left[\frac{b^5}{5}\right]$

$I_y=\frac{1}{5}kh{b}^5$

The formula is $I_x={âˆ\neg }_{\mathrm{Î\copyright }}{y}^2\mathrm{ÒÏ}(x,y)dA$

Solving as same in above step

$I_x={∫}_0^b{∫}_0^h{y}^2\mathrm{ÒÏ}(x,y)dydx$

$I_x={∫}_0^b{∫}_0^h{y}^{2}k{x}^{2}dydx\left[\mathrm{ÒÏ}(x,y)=k{x}^2\right]$

$I_x=k{∫}_0^b{x}^2{\left[\frac{y^3}{3}\right]}_0^{h}dx$

$I_x=k{∫}_0^b{x}^2\left[\frac{h^3}{3}\right]dx=\frac{1}{3}k{h}^3{∫}_0^b{x}^{2}dx$

$I_x=\frac{1}{3}k{h}^3{\left[\frac{x^3}{3}\right]}_0^b$

Hence,$I_x=\frac{1}{9}k{h}^3{b}^3$

The mass is given by $m={âˆ\neg }_{\mathrm{Î\copyright }}\mathrm{ÒÏ}(x,y)dA$

As $\mathrm{ÒÏ}(x,y)=k{x}^2$

$m={∫}_0^b{∫}_0^{h}k{x}^{2}dydx$

Solving inner integral

$m={∫}_0^{b}k{x}^2[y{]}_0^{h}dx$

$m={∫}_0^{b}k{x}^{2}hdx=kh{∫}_0^b{x}^{2}dx$

Solving further

$m=kh{\left[\frac{x^3}{3}\right]}_0^b$

$⇒m=\frac{1}{3}kh{b}^3$

It is given by

$R_y=\sqrt{\frac{I_y}{m}}\text{and}{R}_x=\sqrt{\frac{I_x}{m}}$

$R_y=\sqrt{\frac{\frac{1}{5}kh{b}^5}{\frac{1}{3}kh{b}^3}}\text{and}{R}_x=\sqrt{\frac{\frac{1}{9}k{h}^3{b}^3}{\frac{1}{3}kh{b}^3}}$

$R_y=\sqrt{\frac{3}{5}}b\text{and}{R}_x=\frac{h}{\sqrt{3}}$

Putting values

$R_y=\sqrt{\frac{147}{50}}\text{and}{R}_x=\sqrt{\frac{147}{150}}$