Answer
$I_x=0.133m^4$
Work Step by Step
We can find the required moment of inertia about the x-axis as follows:
$I_x=\int y^2 dA$.ea(1)
As $y=x^{\frac{1}{2}}$ $\implies x=y^2$
We know that $dA=(1-x)dy$
$\implies dA=(1-y^2)dy$
Thus, the equation (1) becomes $I_x=\int y^2(1-y^2)dy$
$\implies I_x=\int_0^1 y^2(1-y^2)dy$
$\implies I_x=\int_0^1(y^2-y^4)dy$
$\implies I_x=\frac{y^3}{3}|^1_0-\frac{y^5}{5}|_0^1$
After applying the limits, we obtain:
$I_x=\frac{1}{3}-\frac{1}{5}$
$\implies I_x=0.133m^4$
