Centroidal moment of inertia

$\bar{I} = \frac{1}{12}(BH^3 - bh^3)$

$\bar{I} = \frac{1}{12}[ \, 8(8^3) - 4(4^3) \, ]$

$\bar{I} = 320 \, \text{ in.}^4$

$\bar{I}_x = \bar{I}_y = \bar{I}$

$\bar{I}_x = \bar{I}_y = 320 \, \text{ in.}^4$

Area

$A = 8^2 - 4^2$

$A = 48 \, \text{ in.}^2$

By transfer formula for moment of inertia

$I = \bar{I} + Ad^2$

$I = 320 + 48(4^2)$

$I = 1088 \, \text{ in.}^4$

$I_x = I_y = I$

$I_x = I_y = 1088 \, \text{ in.}^4$

Polar moment of inertia

$J = I_z = I_x + I_y$

$J = 1088 + 1088$

$J = 2176 \, \text{ in.}^4$ *answer*

Radius of gyration

$k_z = \sqrt{\dfrac{J}{A}} = \sqrt{\dfrac{2176}{48}}$

$k_z = 6.733 \, \text{ in.}$ *answer*