# Archimedes Proof

• Feb 26th 2008, 06:11 PM
doctorgk
Archimedes Proof
The area of a parabolic arch is equal to $\displaystyle 2/3$ the product of the base and the height.

I need to prove this formula for the general parabola $\displaystyle f(x)=ax^2+h$, where $\displaystyle a$ is any real number less than zero
and a definite integral to represent the requested area

thank you
• Feb 26th 2008, 08:08 PM
Soroban
Hello, doctorgk!

Quote:

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 x-intercepts: .$\displaystyle -ax^2 + h \:=\:0\quad\Rightarrow\quad x \:=\:\pm\sqrt{\frac{h}{a}}$

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: .$\displaystyle A \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\left[3h - a\left(\frac{h}{a}\right)\right] \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\cdot(2h)$

$\displaystyle \text{Therefore: }\;A \;=\;\frac{2}{3}\underbrace{\left(2\sqrt{\frac{h}{ a}}\right)}_{\text{base}}\cdot\underbrace{(h)}_{\t ext{height}}$

• Apr 6th 2009, 05:46 AM
futurehsteach
Where did the 2 before the integral in the area equation come from?
• Feb 4th 2010, 03:20 PM
elija49024
Since the integral was only evaluating from 0 to b, and the parabolic arches area is from -b to b, a 2 is added to the front.