**Example 5**

Find the area enclosed by four-leaved rose r = a cos 2θ.

**Solution**

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 Î¸ 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*

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