Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 538: 16

$I_y=780in ^4$

We can find the required moment of inertia as follows: $dA=(4-y)d_x$ Given that $y^2=x$ $\implies y=x^{0.5}$ $\implies dA=(4-x^{0.5})dx$ Now $I_y=\int x^2 dA$ $\implies I_y=\int_0^{16} x^2(4-x^{\frac{1}{2}})dx$ $\implies I_y=4\frac{x^3}{3}|_0^{16}-\frac{x^{\frac{7}{2}}}{\frac{7}{2}}|_0^{16}$ $\implies I_y=4(\frac{4096}{3})-\frac{2}{7}(16384)$ $\implies I_y=780in ^4$