Answer
$I_{xy}=98.4\times (10^6)mm^4$
Work Step by Step
We can find the product of inertia of the cross-sectional area with respect to the x and y axes as follows:
$I_{xy}=\Sigma I_{xy}$
We plug in the known values to obtain:
$I_{xy}=49.2\times 10^6+0+49.2\times 10^6$
This simplifies to:
$I_{xy}=98.4\times (10^6)mm^4$
