Answer
$ I_{xy}=\frac{a^2b^2}{6}$
Work Step by Step
We can determine the required product of inertia for the shaded area with respect to x and y axes as follows:
We know that
$dI_{xy}=0+ydx\cdot x\frac{y}{2}=\frac{xy^2}{2}dx$
Now $I_{xy}=\int_0^a \frac{x^{(\frac{bx^{1/2}}{a^{1/2}})^2}}{2}dx$
This simplifies to:
$\implies I_{xy}=\frac{a^2b^2}{6}$
