Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 539: 19

$I_x=0.267m^4$

We can find the required inertia as follows: $dA=xdy$ As given that $y^2=1-x$ $\implies x=1-y^2$ $\implies dA=(1-y^2)dy$ Now $I_x=\int y^2 dA$ $\implies I_x=\int_{-1}^1 (1-y^2)dy$ $\implies I_x=\frac{y^3}{3}|_{-1}^{1}-\frac{y^5}{5}|_{-1}^{1}$ After applying the limits, we obtain: $I_x=0.267m^4$