Question D.1: A parabolic semisegment OAB is bounded by the x axis, the y ...

To determine the coordinates \overline{ x } and \overline{ y } of the centroid C (Fig. D-5), use Eqs. (D-3a and b). Begin by selecting an element of area dA in the form of a thin vertical strip of width dx and height y. The area of this differential element is

\bar{x}=\frac{Q_{y}}{A}=\frac{\int x d A}{\int d A}   \quad \bar{y}=\frac{Q_{x}}{A}=\frac{\int y d A}{\int d A}             (D-3a,b)

d A=y d x=h\left(1-\frac{x^{2}}{b^{2}}\right) d x            (b)

Therefore, the area of the parabolic semisegment is

A=\int d A=\int_{0}^{b} h\left(1-\frac{x^{2}}{b^{2}}\right) d x=\frac{2 b h}{3}             (c)

Note that this area is two-thirds of the area of the surrounding rectangle. The first moment of an element of area dA with respect to an axis is obtained by multiplying the area of the element by the distance from its centroid to the axis. Since the x and y coordinates of the centroid of the element shown in Fig. D-5 are x and y/2, respectively, the first moments of the element with respect to the x and y axes are

Q_{x}=\int \frac{y}{2} d A=\int_{0}^{b} \frac{h^{2}}{2}\left(1-\frac{x^{2}}{b^{2}}\right)^{2} d x=\frac{4 b h^{2}}{15}             (d)

Q_{y}=\int x d A=\int_{0}^{b} h x\left(1-\frac{x^{2}}{b^{2}}\right) d x=\frac{b^{2} h}{4}           (e)

in which dA has been substituted from Eq. (b).

Now determine the coordinates of the centroid C:

\bar{x}=\frac{Q_{y}}{A}=\frac{3 b}{8} \quad \bar{y}=\frac{Q_{x}}{A}=\frac{2 h}{5}             (f,g)

These results agree with the formulas listed in Appendix E, Case 17.

Notes: The centroid C of the parabolic semisegment also may be located by taking the element of area dA as a horizontal strip of height dy and width:

x=b \sqrt{1-\frac{y}{h}}            (h)

This expression is obtained by solving Eq. (a) for x in terms of y.