Question

Use a double integral to find the area enclosed by one loop of the four-leaved rose r =cos(2 theta).

Answer 1

Consider the four leaved rose $r=cos(2\theta)$

For $r=0$ , solve $\theta$ as,

$cos(2\theta)=0$

$2\theta=\frac{\pi}{2}$ in $[0,\frac{\pi}{2}]$

$\theta=\frac{\pi}{4}$

The sketch of the curve is as shown in the figure below:

The area enclosed by one loop of the rose is evaluated as,

Area$(A)=2\int_{\theta=0}^{\frac{\pi}{4}}\int_{r=0}^{cos(2\theta)}rdrd\theta$ ....[using symmetry]

$=2\int_{\theta=0}^{\frac{\pi}{4}}[\frac{r^2}{2}]_{r=0}^{cos(2\theta)}d\theta$

$=2(\frac{1}{2})\int_{\theta=0}^{\frac{\pi}{4}}cos^2(2\theta)d\theta$

$=\int_{\theta=0}^{\frac{\pi}{4}}[\frac{1+cos(4\theta)}{2}]d\theta$

$=\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{4}}(1+cos(4\theta))d\theta$

$=\frac{1}{2}[\theta+\frac{sin(4\theta)}{4}]_{0}^{\frac{\pi}{4}}$

$=\frac{1}{2}[\frac{\pi}{4}+\frac{0-0}{4}]$

$=\frac{1}{2}[\frac{\pi}{4}+0]$

$=\frac{\pi}{8}$

Therefore, the area of one loop of the rose is Area$(A)=\frac{\pi}{8}$ square units.