Answer
$I_{xy}=5.06in^4$
Work Step by Step
We can find the product of the inertia of the shaded area with respect to the x and y axes as follows:
$I_{xy}=\int_0^3 xy dA$
$\implies I_{xy}=\int_0^3 (x\times \frac{y}{2})(y dx)$
But $y=\frac{1}{9}x^3$
Thus, the above equation becomes
$I_{xy}=\frac{1}{2}\int_0^3 x(\frac{1}{9}x^3)^2 dx$
$\implies I_{xy}=\frac{1}{162}\int_0^3 x^7 dx$
$\implies I_{xy}=\frac{1}{162}\frac{x^8}{8}|_0^3$
After applying the limits, we obtain:
$I_{xy}=5.06in^4$
