Calculating the area of equilateral triangle using calculus

[Adel Makram]

Homework Statement

Calculating the area of equilateral triangle using calculus.

Homework Equations

The Attempt at a Solution

The area of the triangle is the area of the circle minus 3 times the area of the sector shown in (light blue). So, the target is to calculate the pink area first.
$y=\sqrt{r^2-x^2}$
The pink area is $\int_{0}^{x_0} \sqrt{r^2-x^2} dx$ = $\int_{0}^{x_0} r\sqrt{1-\frac{x^2}{r^2}} dx$ Putting $x=r sin a$ and doing the usual math with integration from 0 to $\pi/6$ led me to;
$r^2\int_{0}^{\pi/6} cos^2 a da$=$\frac{r^2}{2}\int_{0}^{\pi/6} cos (2a+1) da$=$\frac{r^2}{2} \left[\frac{sin2a}{2}+a\right]_0^{\pi/6}$=$\frac{r^2}{2} [\frac{1}{2}\frac{\sqrt 3}{2}+\pi/6]$=$\frac{r^2 \sqrt 3}{8} +\frac{\pi}{12}$
The blue area is then $2 (\frac{\pi r^2}{4}-\frac{r^2 \sqrt 3}{8} -\frac{\pi}{12})$=$\frac{\pi r^2}{2}-\frac{r^2 \sqrt 3}{4} -\frac{\pi}{6})$ and then the area of triangle is $\pi r^2-3(\frac{\pi r^2}{2}-\frac{r^2 \sqrt 3}{4} -\frac{\pi}{6})$
This will not give the correct result of $\frac{3\sqrt3 r^2}{4}$

 

[scottdave]

Since you know the angle a = pi/6, you should be able to calculate the x value where the blue area starts, then just integrate sqrt(r² - x²) from there to r. That gets you the top half of the blue area.

 

[willem2]

It's so much easier to integrate the area of the triangle directly.

 

[scottdave]

It's so much easier to integrate the area of the triangle directly.

That too. With what he has figured, it should be easy to figure the equation of the line for the triangle.