Answer
The required area is equal to $\frac{3 \pi}{2}-4$
Work Step by Step
The area of the region D in the first quadrant enclosed by the cardiod is given by:
$$
D=\left\{(r, \theta) | \quad 0 \leq r \leq 1- cos \theta , \quad 0 \leq \theta \leq \frac{\pi}{2} \right\}
$$
By symmetry, the area of the region is 4 times the area of the region D Therefore, the area of enclosed by both of the cardioids $r=1+\cos \theta $, and $r=1+\cos \theta $ is given by:
$$
\begin{aligned}
4 A(D) &=4 \iint_{D} d A=4 \int_{0}^{\pi / 2} \int_{0}^{1-\cos \theta} r d r d \theta \\
&=4 \int_{0}^{\pi / 2}\left[\frac{1}{2} r^{2}\right]_{r=0}^{r=1-\cos \theta} d \theta \\
&=2 \int_{0}^{\pi / 2}(1-\cos \theta)^{2} d \theta \\
&=2 \int_{0}^{\pi / 2}\left(1-2 \cos \theta+\cos ^{2} \theta\right) d \theta \\
&=2 \int_{0}^{\pi / 2}\left[1-2 \cos \theta+\frac{1}{2}(1+\cos 2 \theta)\right] d \theta \\
&=2\left[\theta-2 \sin \theta+\frac{1}{2} \theta+\frac{1}{4} \sin 2 \theta\right]_{0}^{\pi / 2} \\
&=2\left(\frac{\pi}{2}-2+\frac{\pi}{4}\right) \\
&=\frac{3 \pi}{2}-4
\end{aligned}
$$
So, the required area is equal to $\frac{3 \pi}{2}-4$
