$A = {\displaystyle \frac{1}{2}{\int_{\theta_1}}^{\theta_2}} r^2 \, d\theta$
$A = 4 \times {\displaystyle \frac{1}{2} \int_{0}^{\pi / 2}} (2a \, \cos^2 \theta)^2 \, d\theta$
$A = 8a^2 {\displaystyle \int_{0}^{\pi / 2}} (\cos^2 \theta)^2 \, d\theta$
$A = 8a^2 {\displaystyle \int_{0}^{\pi / 2}} \left[ \frac{1}{2}(1 + \cos 2\theta) \right]^2 \, d\theta$
$A = 2a^2 {\displaystyle \int_{0}^{\pi / 2}} (1 + 2 \cos 2\theta + \cos^2 2\theta) \, d\theta$
$A = 2a^2 {\displaystyle \int_{0}^{\pi / 2}} \left[1 + 2 \cos 2\theta + \frac{1}{2}(1 + \cos 4\theta) \right] \, d\theta$
$A = 2a^2 {\displaystyle \int_{0}^{\pi / 2}} (\frac{3}{2} + 2 \cos 2\theta + \frac{1}{2}\cos 4\theta) \, d\theta$
$A = 2a^2 \left[ \dfrac{3}{2}\theta + \sin 2\theta + \dfrac{\sin 4\theta}{8} \right]_{0}^{\pi / 2}$
$A = 2a^2 \left[ \left( \dfrac{3\pi}{4} + \sin \pi + \dfrac{\sin 2\pi}{8} \right) - \left( 0 + \sin 0 + \dfrac{\sin 0}{8} \right) \right]$
$A = 2a^2 \left[ \left( \frac{3}{4}\pi + 0 + 0 \right) - \left( 0 + \sin 0 + 0 \right) \right]$
$A = \frac{3}{2}\pi a^2$ answer
