- #1

Adel Makram

- 635

- 15

## 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}##