Answer
$I_{xy}=97.8in^4$
Work Step by Step
We can find the product of the moment of inertia of the cross-sectional area with respect to the x and y axes as follows:
$I_{xy,1}=Ix^{\prime}y^{\prime},1+x^{-}y^{-}A$
$I_{xy,1}=0+0.5(4)(8\times 1)=16in^4$
and $I_{xy,2}=I_{x^{\prime}y^{\prime},2}+x^{-}y^{-}A$
$\implies I_{xy,2}=0+6(0.5)(10)(1)=30in^4$
Similarly $I_{xy,3}=I_{x^{\prime}y^{\prime},3}+x^{-}y^{-}A$
$\implies I_{xy,3}=0+11.5(1.5)(3\times 1)=51.75in^4$
Now $I_{xy}=\Sigma I_{xy,i}$
$\implies I_{xy}=16+30+51.75=97.8in^4$
