Engineering Mechanics: Statics & Dynamics (14th Edition)

$I_{xy}=5.06in^4$
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: $I_{xy}=5.06in^4$