$r^2 = 16 \cos \theta$

**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*