Answer
$I_x=1845in^4$
Work Step by Step
We can find the required moment of inertia as follows:
$I_x=I+Ad^2_y$
We plug in the known values to obtain:
$I_x=\frac{(6)(9)^3}{36}+\frac{(6)(9)}{2}(6)^2+\frac{6(3)^3}{36}+\frac{(6)(3)}{2}(7)^2+\frac{6(6)^3}{12}+6(6)(3)^2-(\frac{\pi}{4}(2)^4)+(2)^2\pi (3)^2)$
This simplifies to:
$I_x=1845in^4$
