Answer
$I_{xy}=\frac{a^4}{280}$
Work Step by Step
We can find the product of the moment of inertia with respect to the x and y axes as follows:
$dA=ydx$
$dx=(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$
As $x^{\prime}=x$
$\implies y^{\prime}=\frac{y}{2}$
$\implies y^{\prime}=\frac{1}{2}(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$
Now $dI_{XY}=dI_{xy}+dAx{\prime}y^{\prime}$
$\implies dI_{XY}=0+((a^{\frac{1}{2}}-x^{\frac{1}{2}})^2dx) x(\frac{1}{2})(a^{\frac{1}{2}}-x^{\frac{1}{2}})^2$
$\implies dI_{XY}=\frac{1}{2}(x^3+a^2 x+6ax^2-4a^{\frac{3}{2}}x^{\frac{3}{2}}-4a^{\frac{1}{2}}x^{\frac{5}{2}})$dx
We know that
$I_{xy}=\int dI_{xy}$
$\implies I_{xy}=\frac{1}{2}\int_0^a (x^3+a^2 x+6ax^2-4a^{\frac{3}{2}}x^{\frac{3}{2}}-4a^{\frac{1}{2}}x^{\frac{5}{2}})dx$
This simplifies to:
$I_{xy}=\frac{a^4}{280}$
