Determine the moment of inertia for the shaded area about the x axıs

Write the equation of the shaded area.
${\displaystyle y^2=x}$
${\displaystyle x=y^2}$
Consider a rectangular differential element with respect to the y-axis with a thickness dy and intersects the boundary at (x, y).
Express the area of the differential element parallel to the x-axis.
${\displaystyle dA=xdy}$
Substitute ${\displaystyle y^2}$ for x.
${\displaystyle dA=y^{2}dy}$
Calculate the moment of inertia for the shaded area about the x-axis.$I_x={\int }_A{y}^{2}dA$
Substitute ${\displaystyle y^{2}dy}$
$I_x={\int }_0^{4in.}{y}^2(y^{2}dy)$
${\displaystyle I_x={\int }_0^{4in.}\left(y^{4}dy\right)}$
${\displaystyle I_x=\frac{y^5}{7}{\mid }_0^{4in.}}$
${\displaystyle =\frac{{(4in.)}^5-0}{7}}$
${\displaystyle =204.8i{n}^4}$

at the end, should be 1/5 instead of 1/7

Yes, it should be $\frac{y^5}{5}$