For $r = a \sin 3\theta$
$\displaystyle A = 6 \times \frac{1}{2}\int_{0}^{\pi/6} (a \sin 3\theta)^2 \, d\theta$
$\displaystyle A = 3a^2 \int_{0}^{\pi/6} \sin^2 3\theta \, d\theta$
$A = 3a^2 {\displaystyle \int_{0}^{\pi/6}} \frac{1}{2}(1 - \cos 6\theta) \, d\theta$
$A = \frac{3}{2}a^2 {\displaystyle \int_{0}^{\pi/6}} (1 - \cos 6\theta) \, d\theta$
$A = \frac{3}{2}a^2 \left[ \theta - \dfrac{\sin 6\theta}{6} \right]_{0}^{\pi/6}$
$A = \frac{3}{2}a^2 \left[ \left( \dfrac{\pi}{6} - \dfrac{\sin \pi}{6} \right) - \left( 0 - \dfrac{\sin 0}{6} \right) \right]$
$A = \frac{3}{2}a^2 \left[ \left( \dfrac{\pi}{6} - 0 \right) - \left( 0 - 0 \right) \right]$
$A = \frac{1}{4}\pi a^2$ answer
For $r = a \cos 3\theta$
$\displaystyle A = 6 \times \frac{1}{2}\int_{0}^{\pi/6} (a \cos 3\theta)^2 \, d\theta$
$\displaystyle A = 3a^2 \int_{0}^{\pi/6} \cos^2 3\theta \, d\theta$
$A = 3a^2 {\displaystyle \int_{0}^{\pi/6}} \frac{1}{2}(1 + \cos 6\theta) \, d\theta$
$A = \frac{3}{2}a^2 {\displaystyle \int_{0}^{\pi/6}} (1 + \cos 6\theta) \, d\theta$
$A = \frac{3}{2}a^2 \left[ \theta + \dfrac{\sin 6\theta}{6} \right]_{0}^{\pi/6}$
$A = \frac{3}{2}a^2 \left[ \left( \dfrac{\pi}{6} + \dfrac{\sin \pi}{6} \right) - \left( 0 + \dfrac{\sin 0}{6} \right) \right]$
$A = \frac{3}{2}a^2 \left[ \left( \dfrac{\pi}{6} + 0 \right) - \left( 0 + 0 \right) \right]$
$A = \frac{1}{4}\pi a^2$ answer
