Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 538: 11

$I_x=614m^4$

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: $I_x=614m^4$