Archimedes Proof

**doctorgk** · February 26th 2008, 06:11 PM

The area of a parabolic arch is equal to $2/3$ the product of the base and the height.

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

thank you

**Soroban** · February 26th 2008, 08:08 PM

Hello, doctorgk!

The area of a parabolic arch is equal to $2/3$ the product of the base and the height.

Prove this formula for the general parabola $f(x)\:=\:-ax^2+h$

They could have used a nicer general form . . .

Code:

                |
               h*
             *  |  *
           *    |    *
          *     |     *
                |
      - -*- - - + - - -*- -
         b      |      b

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

We have: . $A \;\;=\;\;2\int^b_0\left(-ax^2 + h\right)\,dx \;\;=\;\;2\left(-\frac{ax^3}{3} + hx\right)\,\bigg]^b_0$

. . $= \;\frac{2}{3}x\left(3h - ax^2\right)\,\bigg]^b_0 \;\;=\;\;\frac{2}{3}\,b\left(3h - ab^2\right)$

Since $b \:=\:\sqrt{\frac{h}{a}}$ , we have: . $A \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\left[3h - a\left(\frac{h}{a}\right)\right] \;=\;\frac{2}{3}\sqrt{\frac{h}{a}}\cdot(2h)$

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

**futurehsteach** · April 6th 2009, 05:46 AM

Where did the 2 before the integral in the area equation come from?

**elija49024** · February 4th 2010, 03:20 PM

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.

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