06 Area Within the Curve r^2 = 16 cos θ

$r^2 = 16 \cos \theta$
 

θ	0°	±30°	±60°	±90°	> 90°
r	±4	±3.72	±2.83	0	imaginary

 

The values in the table show that the graph is symmetrical to the origin and θ ranges from -90° to 90°.
 

 

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

$A = 4\left[ {\displaystyle \frac{1}{2}{\int_0}^{\,90^\circ}} 16 \cos \theta \, d\theta \right]$

$A = 32 \Big[ \sin \theta \Big]_0^{90^\circ}$

$A = 32 \Big[ \sin 90^\circ - \sin 0^\circ \Big]$

$A = 32 \, \text{ units}^2$           answer