Q2 (15 points) Making use of the centroid information for a parabolic spandrel given in Table Q2
below, deduce/derive the formulas for the centroidal location ($\bar{x}_{comp}$, $\bar{y}_{comp}$) of the complementary
shaded area $A_{comp}$ above the parabola in Figure Q2.
Finding the area
$A = ha - \int_{0}^{a} \int_{0}^{kx^2} dy dx$
$A = ha - \int_{0}^{a} kx^2 dx$
$A = ha - k[\frac{x^3}{3}]_{0}^{a}$
$A = ha - \frac{ka^3}{3}$
So $A = ha - \frac{ah}{3}$
$A = \frac{2ah}{3} + 2$
Finding x center centroid
$\bar{x}A = \int_{0}^{a} \int_{0}^{h} x dy dx - \int_{0}^{a} \int_{0}^{kx^2} x dy dx + 2$
$\bar{x}A = \int_{0}^{a} [xy]_{0}^{h} dx - \int_{0}^{a} [xy]_{0}^{kx^2} dx + 2$
$\bar{x}A = \int_{0}^{a} xh dx - \int_{0}^{a} x kx^2 dx + 2$
$\bar{x}A = h [\frac{x^2}{2}]_{0}^{a} - k \int_{0}^{a} x^3 dx + 2$
$\bar{x}A = \frac{ha^2}{2} - k [\frac{x^4}{4}]_{0}^{a} + 2$
$\bar{x}A = \frac{ha^2}{2} - \frac{ka^4}{4} + 2$
$\bar{x}A = \frac{a^2h}{2} - \frac{a^2h}{4}$
So $\bar{x}A = \frac{a^2h}{4}$
$\bar{x} = \frac{a^2h}{4} / \frac{2ah}{3}$
$\bar{x} = \frac{a^2h}{4} \cdot \frac{3}{2ah}$
$\bar{x} = \frac{3a}{8} + 2$
Table Q2: centroidal information
$\bar{x} = \frac{3a}{4}$
$\bar{y} = \frac{3h}{10}$
Area = $\frac{ah}{3}$
