Answer
$I_x=1192.37in^4$
Work Step by Step
We can find the required moment of inertia as follows:
$I_x=\Sigma (I+d^2_y)$
We plug in the known values to obtain:
$\implies I_x=\frac{6(10)^3}{12}+6(10)(5)^2-\frac{3(6)^3}{36}+\frac{6(3)(8)^2}{2}-(\frac{\pi^2}{4}+(2)^2\pi (4)^2)$
This simplifies to:
$I_x=1192.37in^4$
