# Chapter 10 - Moments of Inertia - Section 10.3 - Radius of Gyration of an Area - Problems - Page 538: 15

$I_x=204.8in^4$

#### Work Step by Step

The required moment of inertia can be determined as follows: $dA=xdy$ Given that $y^2=x$ $\implies dA=y^2 dy$ Now $I_x=\int y^2 dA$ $\implies I_x=\int _0^4 y^2 y^2 dy$ $\implies I_x=\int_0^4 y^4dy$ $\implies I_x=\frac{y^5}{5}|_0^4$ $\implies I_x=\frac{1024}{5}$ $\implies I_x=204.8in^4$

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