05 Area Enclosed by Four-Leaved Rose r = a cos 2θ

$r = a \cos 2\theta$
 

θ	0°	15°	30°	45°	60°	75°	90°
r	a	0.87a	0.5a	0	-0.5a	-0.87a	-a

 

Since cos (-2θ) = cos 2θ, the equation remains unchanged when θ is replaced by -θ, the curve is symmetric with respect to the x-axis. The equation remains unchanged when θ is replaced by (180° - θ), since cos 2(π - θ) = cos 2θ. Therefore, the graph is symmetric with respect to the y-axis. Because of symmetry, we can sketch the curve without recourse to point-by-point plotting.
 

 

$A = {\displaystyle \frac{1}{2}{\int_{\theta_1}}^{\theta_2}} r^2 \, d\theta$

$A = 8 \left[ {\displaystyle \frac{1}{2}{\int_0}^{\pi/4}} a^2 \cos^2 2\theta \, d\theta \right]$

$A = {\displaystyle 4a^2{\int_0}^{\pi/4}} \cos^2 2\theta \, d\theta$

$A = {\displaystyle 4a^2{\int_0}^{\pi/4}} \frac{1}{2}(1 + \cos 4\theta) \, d\theta$

$A = {\displaystyle 2a^2{\int_0}^{\pi/4}} (1 + \cos 4\theta) \, d\theta$

$A = 2a^2 \Big[ \theta + \frac{1}{4} \sin 4\theta \Big]_0^{\pi/4}$

$A = 2a^2 \Big[ \frac{1}{4}\pi + \frac{1}{4} \sin \pi \Big] - 2a^2 \Big[ 0 + \frac{1}{4} \sin 0 \Big]$

$A = \frac{1}{2}\pi a^2 \, \text{ unit}^2$           answer