Q. 11 - Calculus

Question

Give a geometric explanation why $n{\int }_0^{2\pi /n}{\int }_0^{R}rdrd\theta =\pi R^2$

for any positive real number $\mathrm{ℝ}$ and any positive integer $n$.

Would the equation also hold for nonintegral values of $n$?

Step-by-Step Solution

Answer

Area of $n$ sectors = $n{\int }_0^{2\pi /n}{\int }_0^{\pi }rdrd\theta =\pi R^2\text{ [Area of a circle] }$ is shown.

Step 1 Given Information

The objective of this problem is to give the geometric explanation why

$n{\int }_0^{2\pi /n}{\int }_0^{\pi }rdrd\theta =\pi R^2$

Step 2 Calculation

Suppose a circle of radius $R$ is divided into $n$ sectors of equal areas. Each sector will subtend an angle of $\frac{2\mathrm{\pi }}{n}$ at the center of circle. Area of a sector can be calculated in polar form as an integral.

Area of a sector $={\int }_0^{2\pi i=}{\int }_0^{R}rdrd\theta$

Integrate with respect to $r$.

${\int }_0^{2\pi /n}{\int }_0^{\pi }rdrd\theta ={\int }_0^{2\pi /n}{\left[\frac{r^2}{2}\right]}_0^{\pi }d\theta$

${\int }_0^{2{\pi }^{\text{'}/n}}{\int }_0^{\pi }rdrd\theta ={\int }_0^{2\sin }\frac{R^2}{2}d\theta$

${\int }_0^{2\pi /n}{\int }_0^{R}rdrd\theta =\frac{R^2}{2}{\int }_0^{2\pi /n}d\theta$

Integrate with respect to $\theta$.

${\int }_0^{2\pi /n}{\int }_0^{n}rdrd\theta =\frac{R^2}{2}[\theta {]}_0^{2\pi /m}$

Substitute the limits

${\int }_0^{2\pi /n}{\int }_0^{\pi }rdrd\theta =\frac{\pi R^2}{n}$

Therefore, area of $n$ sectors

$n{\int }_0^{2\sin }{\int }_0^{R}rdrd\theta =\frac{\pi n{R}^2}{n}$

$n{\int }_0^{2\pi /n}{\int }_0^{\pi }rdrd\theta =\pi R^2\text{ [Area of a circle] }$