Question D.3: Determine the moments of inertia Ix and I y for the paraboli...
Determine the moments of inertia I_{x} and I_{y} for the parabolic semisegment OAB shown in Fig. D-12. The equation of the parabolic boundary is
y=f(x)=h\left(1-\frac{x^{2}}{b^{2}}\right) (a)
(This same area was considered previously in Example D-1.)
To determine the moments of inertia by integration, use Eqs. (D-9a and b). The differential element of area dA is selected as a vertical strip of width dx and height y, as shown in Fig. D-12. The area of this element is
I_{x}=\int y^{2} d A \quad I_{y}=\int x^{2} d A (D-9a,b)
d A=y d x=h\left(1-\frac{x^{2}}{b^{2}}\right) d x (b)
Since every point in this element is at the same distance from the y axis, the moment of inertia of the element with respect to the y axis is x^{2}dA. Therefore, the moment of inertia of the entire area with respect to the y axis is obtained as
I_{y}=\int x^{2} d A=\int_{0}^{b} x^{2} h\left(1-\frac{x^{2}}{b^{2}}\right) d x=\frac{2 h b^{3}}{15} (c)
To obtain the moment of inertia with respect to the x axis, note that the differential element of area dA has a moment of inertia dI_{x} with respect to the x axis equal to
d I_{x}=\frac{1}{3}(d x) y^{3}=\frac{y^{3}}{3} d xas obtained from Eq. (D-12). Hence, the moment of inertia of the entire area with respect to the x axis is
I_{B B}=\int y^{2} d A=\int_{0}^{h} y^{2} b d y=\frac{b h^{3}}{3} (D-12)
I_{x}=\int_{0}^{b} \frac{y^{3}}{3} d x=\int_{0}^{b} \frac{h^{3}}{3}\left(1-\frac{x^{2}}{b^{2}}\right)^{3} d x=\frac{16 b h^{3}}{105} (d)
These same results for I_{x} and I_{y} can be obtained by using an element in the form of a horizontal strip of area dA = xdy or by using a rectangular element of area dA = dxdy and performing a double integration. Also, note that the preceding formulas for I_{x} and I_{y} agree with those given in Case 17 of Appendix E.