Answer
$I_x=\frac{2}{15}bh^3$
Work Step by Step
We can find the required moment of inertia about the x-axis as follows:
As given that $y^2=\frac{h^2}{b}x$
$\implies x=\frac{by^2}{h^2}$
We known that
$dA=(b-x)dy$
$\implies dA=(b-\frac{by^2}{h^2})dy$
Now $I_x=\int_0^h y^2 dA$
$\implies I_x=\int_0^h y^2 (b-\frac{by^2}{h^2})dy$
$\implies I_x=\int_0^h y^2 b(1-\frac{y^2}{h^2})dy$
After taking the integration and applying the limits, we obtain:
$I_x=b(\frac{y^3}{3}-\frac{y^5}{5h^2})$
$\implies I_x=\frac{2}{15}bh^3$
