Answer
$\dfrac{3\pi a^2}{2}$
Work Step by Step
Consider $A=\dfrac{1}{2}\int_m^n (r^2) d \theta =\int_{0}^{2\pi} \dfrac{a(1+ \cos \theta)^2}{2} d \theta$
$\int_{0}^{2\pi} \dfrac{a(1+ \cos \theta)^2}{2} d \theta=\dfrac{a^2}{2}\int_{0}^{2\pi} 1+(\dfrac{1}{2})+\dfrac{\cos2 \theta}{2}+2\cos \theta d \theta$
or, $\dfrac{a^2}{2}\int_{0}^{2\pi} 1+(\dfrac{1}{2})+\dfrac{\cos2 \theta}{2}+2 (\cos \theta) d \theta=\dfrac{a^2}{2}[ 3\theta+\dfrac{\sin (2 \theta) }{4}+(2) \sin \theta]_{0}^{2\pi} $
Thus, $A=(\dfrac{a^2}{2}) [\dfrac{6\pi}{2}]=(\dfrac{3\pi}{2}) a^2$
