Q.63 - Calculus

Question

 Use a double integral to prove that the area of the circle with radius R and equation r=2$R\cos \theta is\pi R^2.$

Step-by-Step Solution

Answer

The areal of the circle is

$A=\pi R^2$

part (a) step 1: Given informational

The objective of this problem is to use double integral to prove that the area of the circle with radius R and equation $r=2R\cos \theta$ is $\pi R^2.$

part (b) step 2: Draw the circle

part (c) step 3: calculation

Plot of$r=2R\cos \theta$

Given circle is symmetrical about the horizontal axis. Therefore area of circle in polar form can be expressed as the twice of area of upper half circle.

$A=2{\int }_a^{\infty }{\int }_5^{2}rdrd\theta$

Here,

${\theta }_1=0,{\theta }_2=\frac{\pi }{2}and{r}_1=0,r_2=r$

$A=2{\int }_0^{\pi /2}{\int }_0^{r-2ton\theta }rdrd\theta$

Integrate with respect to $r$ first

$A=2{\int }_0^{\pi /2}{\left[\frac{r^2}{2}\right]}_0^{2RRene}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

$A=2{\int }_0^{\pi /2}\left[\frac{(2R\cos \theta {)}^2-0}{2}\right]$

$A=2R^2{\int }_0^{z/2}\left[2{\cos }^2\theta \right]d\theta$

A=2 R^{2} \int_{0}^{2 / 2}[1+\cos 2 \theta] \theta

Inteorate with respect to \theta

&A=2 R^{2}\left[\theta+\frac{1}{2} \sin 2 \theta\right]_{0}^{x / 2}\left[\int \cos x d x=\sin x+C\right] \\

&A=2 R^{2}\left[\frac{\pi}{2}+\frac{1}{2} \sin \pi-0\right] \\

&A=\pi R^{2}