#### Answer

$I_x=614m^4$

#### Work Step by Step

We can find the required moment of inertia as follows:
Given that $y=\frac{1}{8}x^3$
$\implies x=2y^{\frac{1}{3}}$
We know that
$dA=xdy$
$\implies dA=2y^{\frac{1}{3}}dy$
Now $I_x=\int_0^8 y^2(2y^{\frac{1}{3}})dy$
$\implies I_x=2\int_0^8 y^{\frac{7}{3}}dy $
$\implies I_x=2(\frac{3}{10}y^{\frac{10}{3}})|_0^8$
After applying the limits, we obtain:
$I_x=614m^4$