Answer
$I_{max}=309in^4$
$I_{min}=42.1in^4$
$\theta_1=-31.4^{\circ}$
$\theta_2=58.6^{\circ}$
Work Step by Step
We can determine the directions of the principal axes and the principal moments of inertia as follows:
$I_x=\Sigma (I+Ad^2_y)$
$\implies I_x=\frac{4(6)^3}{12}+4(6)(3)^2-\frac{\pi (1)^4}{4}-(1)^2\pi(4)^2$
$\implies I_x=236.95in^4$
The moment of inertia about the y-axis is given as
$I_y=\Sigma (I+Ad^2_x)$
$\implies I_y=\frac{(6)(4)^3}{12}+(4)(6)(2)^2-\frac{\pi(1)^4}{4}-(1)^2\pi (2)^2=114.65in^4$
The product of inertia about the x and y axes is given as
$I_{xy}=\Sigma (I+Ad_x d_y)$
$\implies I_{xy}=0+4(6)(3)(2)-0-(1)^2\pi (4)(2)=118.87in^4$
Now the minimum and maximum moment of inertia can be calculated as
$I_{max}=\frac{I_x+I_y}{2}+\sqrt{(\frac{I_x-I_y}{2})^2+I^2_{xy}}$
We plug in the known values to obtain:
$I_{max}=\frac{236.95+114.65}{2}-\sqrt{(\frac{236.95-114.95}{2})^2+(118.87)^2}$
$\implies I_{max}=309in^4$
and the minimum moment of inertia is given as
$I_{min}=\frac{I_x+I_y}{2}-\sqrt{(\frac{I_x-I_y}{2})^2+I^2_{xy}}$
We plug in the known values to obtain:
$I_{min}=\frac{236.95+114.65}{2}-\sqrt{(\frac{236.95-114.95}{2})^2+(118.87)^2}$
$\implies I_{max}=42.1in^4$
