Question 12.4: The parabolic semisegment OAB shown in Fig. 12-15 has base b......

We can use the parallel-axis theorem (rather than integration) to find the centroidal moments of inertia because we already know the area A. the centroidal coordinates $\bar{x}$ and $\bar{y}$. and the moments of inertia $I_x$ and $I_y$ with respect to the x and y axes. These quantities were obtained earlier in Examples 12-1 and 12-3. They also are listed in Case 17 of Appendix D and are repeated here:

$A=\frac{2bh}{3}\quad \bar{x}=\frac{3b}{8}\quad \bar{y}=\frac{2h}{5}\quad I_x=\frac{16bh^3}{105}\quad I_y=\frac{2hb^3}{15}$

To obtain the moment of inertia with respect to the $x_c$ axis, we write the parallel-axis theorem as follows:

$I_x=I_{x_c}+A\bar{y}^2$

Therefore, the moment of inertia $I_{x_c}$ is

$I_{x_c}=I_x\ -\ A\bar{y}^2=\frac{16bh^3}{105}\ -\ \frac{2bh}{3}\left(\frac{2h}{5}\right)^2=\frac{8bh^3}{175}$                 (12-13a)

In a similar manner, we obtain the moment of inertia with respect to the $y_c$ axis:

$I_{x_c}=I_y\ -\ A\bar{x}^2=\frac{2hb^3}{15}\ -\ \frac{2bh}{3}\left(\frac{3b}{8}\right)^2=\frac{19bhb^3}{480}$                 (12-13b)

Thus, we have found the centroidal moments of inertia of the semisegment.