Answer
$I_x=0.267m^4$
Work Step by Step
We can find the moment of inertia about the x-axis as follows:
$dA=xdy$
As given that $y^2=1-0.5x$
$\implies x=2(1-y^2)$
$\implies dA=2(1-y^2)dy$
Now $I_x=\int y^2 dA$
$\implies I_x=\int_0^1 y^2 \cdot 2(1-y^2)dy$
$\implies I_x=2\int_0^1 (y^2-y^4)dy$
$\implies I_x=2(\frac{y^3}{3}|_0^1-\frac{y^5}{5}|_0^1)$
After applying the limits, we obtain:
$I_x=2(\frac{1}{3}-\frac{1}{5})$
This simplifies to:
$I_x=0.267m^4$
