Hello, doctorgk!
The area of a parabolic arch is equal to $\displaystyle 2/3$ the product of the base and the height.
Prove this formula for the general parabola $\displaystyle f(x)\:=\:ax^2+h$ They could have used a nicer general form . . . Code:

h*
*  *
*  *
*  *

 *   +   * 
b  b
For the xintercepts: .$\displaystyle ax^2 + h \:=\:0\quad\Rightarrow\quad x \:=\:\pm\sqrt{\frac{h}{a}}$
We have: .$\displaystyle A \;\;=\;\;2\int^b_0\left(ax^2 + h\right)\,dx \;\;=\;\;2\left(\frac{ax^3}{3} + hx\right)\,\bigg]^b_0$
. . $\displaystyle = \;\frac{2}{3}x\left(3h  ax^2\right)\,\bigg]^b_0 \;\;=\;\;\frac{2}{3}\,b\left(3h  ab^2\right)$
Since $\displaystyle b \:=\:\sqrt{\frac{h}{a}}$, we have: .$\displaystyle A \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\left[3h  a\left(\frac{h}{a}\right)\right] \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\cdot(2h) $
$\displaystyle \text{Therefore: }\;A \;=\;\frac{2}{3}\underbrace{\left(2\sqrt{\frac{h}{ a}}\right)}_{\text{base}}\cdot\underbrace{(h)}_{\t ext{height}}$