Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 538: 12

$I_y=85.3m^4$

We can find the required moment of inertia as follows: $dA=(8-y)dx$ $\implies dA=(8-\frac{1}{8}x^3)dx$ Now $I_y=\int_0^4 x^2 dA$ $\implies I_y=\int_0^4 x^2 (8-\frac{1}{8}x^3)dx$ $\implies I_y=8\int_0^4x^2 dx-\frac{1}{8}\int_0^4 x^5 dx$ $\implies I_y=8\frac{x^3}{3}|_0^4-\frac{1}{8}\frac{x^6}{6}|_0^4$ This simplifies to: $I_y=85.3m^4$