91影视

The objective of this problem is to use double integral to prove that the area of the circle with radius R and equation $r=2R\cos 胃$ isrole="math" localid="1650465631819" $蟺R^2.$

Plot of$r=2R\cos 胃$

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.

$A=2{鈭�}_a^{鈭�}{鈭�}_5^{2}rdrd胃$

Here,

Integrate with respect to $r$ first

$A=2{鈭�}_0^{蟺/2}{\left[\frac{r^2}{2}\right]}_0^{2RRene}d胃\left[鈭�x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

$A=2{鈭�}_0^{蟺/2}\left[\frac{(2R\cos 胃{)}^2-0}{2}\right]$

$A=2R^2{鈭�}_0^{z/2}\left[2{\cos }^2胃\right]d胃$

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}