Answer
$$2+\dfrac{\pi}{4}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ A=2\int^{\pi/2}_0 \int ^{1+cos\theta}_1 \space r \space dr \space d\theta \\ =2\int^{\pi/2}_0 (r^2/2) ^{1+cos\theta}_1 \space d\theta \\= 2 \int^{\pi/2}_0 \cos\theta + \int^{\pi/2}_0 \dfrac{1+\cos 2\theta}{ 2} d \theta \\ = \int^{\pi/2}_0 (2cos\theta+cos^2\theta) d\theta \\=\dfrac{8+\pi}{4} \\= 2+\dfrac{\pi}{4}$$
